We remember the happy school years. Pioneers in the lessons of mathematics, starting to study the roots, first of all got acquainted with the square root. We will go the same way.
Example 1
Find the indefinite integral
Analyzing the integrand, one comes to the sad conclusion that it does not at all resemble tabular integrals. Now, if all this goodness was in the numerator, it would be easy. Or there would be no root at the bottom. Or a polynomial. None fraction integration methods don't help either. What to do?
The main technique for solving irrational integrals is the replacement of a variable, which will save us from ALL roots in the integrand.
Note that this replacement is a bit peculiar, its technical implementation differs from the “classical” replacement method, which was discussed in the lesson. Replacement method in indefinite integral.
In this example, you need to replace x = t 2 , that is, instead of "x" under the root we will have t 2. Why is this a replacement? Because, and as a result of the replacement, the root will disappear.
If we had instead of the square root in the integrand, then we would have made a replacement. If there was, then they would have held and so on.
Okay, we will turn into . What will happen to the polynomial? There are no difficulties: if , then 
.
It remains to find out what the differential will turn into. It is done like this:
We take our replacement and hang differentials on both parts:
(let's write in as much detail as possible).
The solution design should look something like this:
.
Let's replace: 
.
.
(1) We carry out the substitution after the replacement (how, what and where, has already been considered).
(2) We take the constant outside the integral. The numerator and denominator are reduced by t.
(3) The resulting integral is tabular, we prepare it for integration by selecting a square.
(4) We integrate over the table using the formula
.
(5) We carry out the reverse replacement. How it's done? We remember why they danced: if, then.
Example 2
Find the indefinite integral
This is a do-it-yourself example. Full solution and answer at the end of the lesson.
Somehow it happened that in Examples 1, 2 there is a “naked” numerator with a single differential . Let's fix the situation.
Example 3
Find the indefinite integral
A preliminary analysis of the integrand again shows that there is no easy way. And so you need to get rid of the root.
Let's replace: .
Behind denote the whole expression under the root. The replacement from the previous examples is not suitable here (more precisely, it can be done, but this will not save us from the root).
We hang differentials on both parts:
Dealt with the numerator. What to do with in the denominator?
We take our replacement and express from it: .
If , then .
(1) We carry out the substitution in accordance with the replacement performed.
(2) Comb the numerator. Here I preferred not to take out the constant outside the integral sign (you can do it this way, it won’t be a mistake)
(3) Expand the numerator into a sum. Once again, we strongly recommend that you read the first paragraph of the lesson. Integration of some fractions. There will be plenty of rigmarole with the expansion of the numerator into a sum in irrational integrals, it is very important to work out this technique.
(4) Divide the numerator by the denominator term by term.
(5) We use the linearity properties of the indefinite integral. In the second integral, we select a square for subsequent integration over the table.
(6) We integrate over the table. The first integral is quite simple, in the second we use the tabular formula of the high logarithm 
.
(7) We carry out the reverse replacement. If we carried out the replacement, then, back: .
Example 4
Find the indefinite integral
This is an example for an independent decision, if you inattentively worked through the previous examples, then make a mistake! Full solution and answer at the end of the lesson.
In principle, integrals with several the same roots, for example
Etc. But what to do if the roots in the integrand different?
Example 5
Find the indefinite integral
So came the retribution for the bare numerators. When such an integral is encountered, it usually becomes scary. But fears are in vain, after carrying out a suitable replacement, the integrand becomes simpler. The task is as follows: to carry out a successful replacement in order to immediately get rid of ALL the roots.
When different roots are given, it is convenient to stick to a certain solution scheme.
First, we write out the integrand on the draft, while representing all the roots in the form:
We will be interested denominators degrees:
There is no universal way to solve irrational equations, since their class differs in number. The article will highlight the characteristic types of equations with substitution using the integration method.
To use the direct integration method, it is necessary to calculate indefinite integrals of the type ∫ k x + b p d x , where p is a rational fraction, k and b are real coefficients.
Example 1
Find and calculate antiderivative functions y = 1 3 x - 1 3 .
Solution
According to the integration rule, it is necessary to apply the formula ∫ f (k x + b) d x \u003d 1 k F (k x + b) + C, and the table of antiderivatives indicates that there is a ready-made solution to this function. We get that
∫ d x 3 x - 1 3 = ∫ (3 x - 1) - 1 3 d x = 1 3 1 - 1 3 + 1 (3 x - 1) - 1 3 + 1 + C = = 1 2 (3 x - 1) 2 3 + C
Answer:∫ d x 3 x - 1 3 = 1 2 (3 x - 1) 2 3 + C .
There are cases where you can use the method of subsuming under the sign of the differential. This is solved by the principle of finding indefinite integrals of the form ∫ f "(x) (f (x)) p d x, when the value of p is considered a rational fraction.
Example 2
Find the indefinite integral ∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x .
Solution
Note that d x 3 + 5 x - 7 \u003d x 3 + 5 x - 7 "d x \u003d (3 x 2 + 5) d x. Then it is necessary to bring under the differential sign using tables of antiderivatives. We get that
∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = ∫ (x 3 + 5 x - 7) - 7 6 (3 x 2 + 5) d x = = ∫ (x 3 + 5 x - 7 ) - 7 6 d (x 3 + 5 x - 7) = x 3 + 5 x - 7 = z = = ∫ z - 7 6 d z = 1 - 7 6 + 1 z - 7 6 + 1 + C = - 6 z - 1 6 + C = z = x 3 + 5 x - 7 = - 6 (x 3 + 5 x - 7) 6 + C
Answer:∫ 3 x 2 + 5 x 3 + 5 x - 7 7 6 d x = - 6 (x 3 + 5 x - 7) 6 + C .
The solution of indefinite integrals provides a formula of the form ∫ d x x 2 + p x + q , where p and q are real coefficients. Then it is necessary to select a full square from under the root. We get that
x 2 + p x + q = x 2 + p x + p 2 2 - p 2 2 + q = x + p 2 2 + 4 q - p 2 4
Applying the formula located in the table of indefinite integrals, we obtain:
∫ d x x 2 ± α = log x + x 2 ± α + C
Then the integral is calculated:
∫ d x x 2 + p x + q = ∫ d x x + p 2 2 + 4 q - p 2 4 = = ln x + p 2 + x + p 2 2 + 4 q - p 2 4 + C = = ln x + p 2 + x 2 + p x + q + C
Example 3
Find an indefinite integral of the form ∫ d x 2 x 2 + 3 x - 1 .
Solution
To calculate, you need to take out the number 2 and place it in front of the radical:
∫ d x 2 x 2 + 3 x - 1 = ∫ d x 2 x 2 + 3 2 x - 1 2 = 1 2 ∫ d x x 2 + 3 2 x - 1 2
Make a selection of the full square in the radical expression. We get that
x 2 + 3 2 x - 1 2 = x 2 + 3 2 x + 3 4 2 - 3 4 2 - 1 2 = x + 3 4 2 - 17 16
Then we get an indefinite integral of the form
Answer: d x x 2 + 3 x - 1 = 1 2 ln x + 3 4 + x 2 + 3 2 x - 1 2 + C
Integration of irrational functions is done in a similar way. Applicable for functions of the form y = 1 - x 2 + p x + q .
Example 4
Find the indefinite integral ∫ d x - x 2 + 4 x + 5 .
Solution
First you need to derive the square of the denominator of the expression from under the root.
∫ d x - x 2 + 4 x + 5 = ∫ d x - x 2 - 4 x - 5 = = ∫ d x - x 2 - 4 x + 4 - 4 - 5 = ∫ d x - x - 2 2 - 9 = ∫ d x - (x - 2) 2 + 9
The table integral looks like ∫ d x a 2 - x 2 = a r c sin x a + C, then we get that ∫ d x - x 2 + 4 x + 5 = ∫ d x - (x - 2) 2 + 9 = a r c sin x - 2 3 +C
Answer:∫ d x - x 2 + 4 x + 5 = a r c sin x - 2 3 + C .
The process of finding antiderivative irrational functions of the form y \u003d M x + N x 2 + p x + q, where the available M, N, p, q are real coefficients, and are similar to the integration of the simplest fractions of the third type. This transformation has several steps:
summing the differential under the root, highlighting the full square of the expression under the root, using tabular formulas.
Example 5
Find antiderivative functions y = x + 2 x 2 - 3 x + 1 .
Solution
From the condition we have that d (x 2 - 3 x + 1) \u003d (2 x - 3) d x and x + 2 \u003d 1 2 (2 x - 3) + 7 2, then (x + 2) d x \u003d 1 2 (2 x - 3) + 7 2 d x = 1 2 d (x 2 - 3 x + 1) + 7 2 d x .
Calculate the integral: ∫ x + 2 x 2 - 3 x + 1 d x = 1 2 ∫ d (x 2 - 3 x + 1) x 2 - 3 x + 1 + 7 2 ∫ d x x 2 - 3 x + 1 = = 1 2 ∫ (x 2 - 3 x + 1) - 1 2 d (x 2 - 3 x + 1) + 7 2 ∫ d x x - 3 2 2 - 5 4 = = 1 2 1 - 1 2 + 1 x 2 - 3 x + 1 - 1 2 + 1 + 7 2 ln x - 3 2 + x - 3 2 - 5 4 + C = = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C
Answer:∫ x + 2 x 2 - 3 x + 1 d x = x 2 - 3 x + 1 + 7 2 ln x - 3 2 + x 2 - 3 x + 1 + C .
The search for indefinite integrals of the function ∫ x m (a + b x n) p d x is carried out using the substitution method.
For the solution it is necessary to introduce new variables:
- When the number p is an integer, then consider that x = z N , and N is the common denominator for m , n .
- When m + 1 n is an integer, then a + b x n = z N , and N is the denominator of p .
- When m + 1 n + p is an integer, then a x - n + b = z N is required, and N is the denominator of p .
Find the definite integral ∫ 1 x 2 x - 9 d x .
Solution
We get that ∫ 1 x 2 x - 9 d x = ∫ x - 1 (- 9 + 2 x 1) - 1 2 d x . It follows that m = - 1 , n = 1 , p = - 1 2 , then m + 1 n = - 1 + 1 1 = 0 is an integer. You can introduce a new variable like - 9 + 2 x = z 2 . It is necessary to express x through z . At the output, we get
9 + 2 x = z 2 ⇒ x = z 2 + 9 2 ⇒ d x = z 2 + 9 2 "d z = z d z - 9 + 2 x = z
It is necessary to perform a substitution into the given integral. We have that
∫ d x x 2 x - 9 = ∫ z d z z 2 + 9 2 z = 2
Answer:∫ d x x 2 x - 9 = 2 3 a r c c t g 2 x - 9 3 + C .
To simplify the solution of irrational equations, the main integration methods are used.
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Complex integrals
This article completes the topic of indefinite integrals, and it includes integrals that I consider quite difficult. The lesson was created at the repeated request of visitors who expressed their wish that more difficult examples be analyzed on the site.
It is assumed that the reader of this text is well prepared and knows how to apply the basic techniques of integration. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Solution examples where you can learn the topic almost from scratch. More experienced students can get acquainted with the techniques and methods of integration, which have not yet been encountered in my articles.
What integrals will be considered?
First, we consider integrals with roots, for the solution of which we successively use variable substitution And integration by parts. That is, in one example, two methods are combined at once. And even more.
Then we will get acquainted with an interesting and original method of reducing the integral to itself. Not so few integrals are solved in this way.
The third number of the program will be integrals of complex fractions, which flew past the cash register in previous articles.
Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid the time-consuming universal trigonometric substitution.
(2) In the integrand, we divide the numerator by the denominator term by term.
(3) We use the property of linearity of the indefinite integral. In the last integral, immediately bring the function under the sign of the differential.
(4) We take the remaining integrals. Note that you can use brackets in the logarithm and not the modulus, because .
(5) We carry out the reverse substitution, expressing from the direct substitution "te":
Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)
As you can see, in the course of the solution, even more than two solution methods had to be used, so to deal with such integrals, you need confident integration skills and not the least experience.
In practice, of course, the square root is more common, here are three examples for an independent solution:
Example 2
Find the indefinite integral
Example 3
Find the indefinite integral
Example 4
Find the indefinite integral
These examples are of the same type, so the complete solution at the end of the article will be only for Example 2, in Examples 3-4 - one answer. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose the same type of examples? Often found in their roles. More often, perhaps, just something like 
.
But not always, when the root of a linear function is under the arc tangent, sine, cosine, exponent, and other functions, several methods have to be applied at once. In a number of cases, it is possible to “get off easy”, that is, immediately after the replacement, a simple integral is obtained, which is taken elementarily. The easiest of the tasks proposed above is Example 4, in which, after the replacement, a relatively simple integral is obtained.
The method of reducing the integral to itself
Clever and beautiful method. Let's take a look at the classics of the genre:
Example 5
Find the indefinite integral
There is a square binomial under the root, and when trying to integrate this example, the teapot can suffer for hours. Such an integral is taken by parts and reduces to itself. In principle, it is not difficult. If you know how.
Let us denote the considered integral by a Latin letter and start the solution: 
Integrating by parts: 
(1) We prepare the integrand for term-by-term division.
(2) We divide the integrand term by term. Perhaps not everyone understands, I will write in more detail:
(3) We use the property of linearity of the indefinite integral.
(4) We take the last integral ("long" logarithm).
Now let's look at the very beginning of the solution: 
And for the ending: 
What happened? As a result of our manipulations, the integral has reduced to itself!
Equate the beginning and end: 
We transfer to the left side with a change of sign: 
And we demolish the deuce to the right side. As a result: 
The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what is the severity here:
Note:
More strictly, the final stage of the solution looks like this:
Thus:
The constant can be re-named with . Why can you rename? Because it still takes any values, and in this sense there is no difference between constants and.
As a result:
A similar trick with constant renaming is widely used in differential equations. And there I will be strict. And here such liberties are allowed by me only in order not to confuse you with unnecessary things and focus on the very method of integration.
Example 6
Find the indefinite integral
Another typical integral for independent solution. Full solution and answer at the end of the lesson. The difference with the answer of the previous example will be!
If there is a square trinomial under the square root, then the solution in any case reduces to the two analyzed examples.
For example, consider the integral 
. All you need to do is in advance select a full square:
.
Next, a linear replacement is carried out, which manages "without any consequences":
, resulting in an integral . Something familiar, right?
Or this example, with a square binomial:
Selecting a full square:
And, after a linear replacement , we get the integral , which is also solved by the already considered algorithm.
Consider two more typical examples of how to reduce an integral to itself:
is the integral of the exponent multiplied by the sine;
is the integral of the exponent multiplied by the cosine.
In the listed integrals by parts, you will have to integrate twice already:
Example 7
Find the indefinite integral
The integrand is the exponent multiplied by the sine.
We integrate by parts twice and reduce the integral to itself: 
As a result of double integration by parts, the integral is reduced to itself. Equate the beginning and end of the solution:
We transfer to the left side with a change of sign and express our integral: 
Ready. Along the way, it is desirable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order.
Now let's go back to the beginning of the example, or rather, to integration by parts: 
For we have designated the exhibitor. The question arises, it is the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what to denote for, one could go the other way: 
Why is this possible? Because the exponent turns into itself (when differentiating and integrating), the sine and cosine mutually turn into each other (again, both when differentiating and integrating).
That is, the trigonometric function can be denoted as well. But, in the considered example, this is less rational, since fractions will appear. If you wish, you can try to solve this example in the second way, the answers must be the same.
Example 8
Find the indefinite integral
This is a do-it-yourself example. Before deciding, think about what is more profitable in this case to designate for, exponential or trigonometric function? Full solution and answer at the end of the lesson.
And, of course, don't forget that most of the answers in this lesson are fairly easy to check by differentiation!
The examples were considered not the most difficult. In practice, integrals are more common, where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will have to get confused in such an integral, and I myself often get confused. The fact is that in the solution there is a high probability of the appearance of fractions, and it is very easy to lose something due to inattention. In addition, there is a high probability of error in signs, note that there is a minus sign in the exponent, and this introduces additional difficulty.
At the final stage, it often turns out something like this:
Even at the end of the solution, you should be extremely careful and correctly deal with fractions: 
Integration of complex fractions
We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, just for one reason or another, the examples were a little “off topic” in other articles.
Continuing the theme of roots
Example 9
Find the indefinite integral
In the denominator under the root there is a square trinomial plus outside the root "appendage" in the form of "X". An integral of this form is solved using a standard substitution.
We decide: 
The replacement here is simple: 
Looking at life after replacement: 
(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) We reduce the numerator and denominator by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integration of some fractions, is solved full square selection method. Select a full square.
(5) By integration, we obtain an ordinary "long" logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at combing the result: under the root, we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral 
This is a do-it-yourself example. Here, a constant is added to the lone x, and the replacement is almost the same: 
The only thing that needs to be done additionally is to express the "x" from the replacement: 
Full solution and answer at the end of the lesson.
Sometimes in such an integral there may be a square binomial under the root, this does not change the way the solution is solved, it will even be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was considered in the lesson Integrals of irrational functions.
Integral of an indecomposable polynomial of the 2nd degree to the degree
(polynomial in denominator)
A rarer, but, nevertheless, occurring in practical examples form of the integral.
Example 13
Find the indefinite integral
But let's go back to the example with the lucky number 13 (honestly, I didn't guess). This integral is also from the category of those with which you can pretty much suffer if you don’t know how to solve.
The solution starts with an artificial transformation: 
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken in parts: 
For an integral of the form ( is a natural number), we have derived recurrent downgrading formula:
, Where 
is an integral of a lower degree.
Let us verify the validity of this formula for the solved integral .
In this case: , , we use the formula: 
As you can see, the answers are the same.
Example 14
Find the indefinite integral
This is a do-it-yourself example. The sample solution uses the above formula twice in succession.
If under the degree is indecomposable square trinomial, then the solution is reduced to a binomial by extracting the full square, for example:
What if there is an additional polynomial in the numerator? In this case, the method of indeterminate coefficients is used, and the integrand is expanded into a sum of fractions. But in my practice of such an example never met, so I skipped this case in the article Integrals of a fractional rational function, I'll skip it now. If such an integral still occurs, see the textbook - everything is simple there. I do not consider it expedient to include material (even simple), the probability of meeting with which tends to zero.
Integration of complex trigonometric functions
The adjective "difficult" for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the methods used to solve the tangent and cotangent are almost the same, so I will talk more about the tangent, meaning that the demonstrated method of solving the integral is also valid for the cotangent.
In the above lesson, we looked at universal trigonometric substitution for solving a certain type of integrals of trigonometric functions. The disadvantage of the universal trigonometric substitution is that its application often leads to cumbersome integrals with difficult calculations. And in some cases, the universal trigonometric substitution can be avoided!
Consider another canonical example, the integral of unity divided by the sine:
Example 17
Find the indefinite integral
Here you can use the universal trigonometric substitution and get the answer, but there is a more rational way. I will provide a complete solution with comments for each step:
(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: In the denominator we divide and multiply by .
(3) According to the well-known formula in the denominator, we turn the fraction into a tangent.
(4) We bring the function under the sign of the differential.
(5) We take the integral.
A couple of simple examples to solve on your own:
Example 18
Find the indefinite integral
Hint: The very first step is to use the reduction formula 
and carefully carry out actions similar to the previous example.
Example 19
Find the indefinite integral
Well, this is a very simple example.
Complete solutions and answers at the end of the lesson.
I think now no one will have problems with integrals: 
and so on.
What is the idea behind the method? The idea is to use transformations, trigonometric formulas to organize only tangents and the derivative of the tangent in the integrand. That is, we are talking about replacing: 
. In Examples 17-19, we actually used this replacement, but the integrals were so simple that it was done with an equivalent action - bringing the function under the differential sign.
Similar reasoning, as I have already mentioned, can be carried out for the cotangent.
There is also a formal prerequisite for applying the above substitution:
The sum of the powers of cosine and sine is a negative integer EVEN number, For example:
for an integral, an integer negative EVEN number.
! Note : if the integrand contains ONLY sine or ONLY cosine, then the integral is taken even with a negative odd degree (the simplest cases are in Examples No. 17, 18).
Consider a couple of more meaningful tasks for this rule:
Example 20
Find the indefinite integral
The sum of the degrees of sine and cosine: 2 - 6 \u003d -4 - a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: 
(1) Let's transform the denominator.
(2) According to the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula 
.
(5) We bring the function under the differential sign.
(6) We carry out the replacement. More experienced students may not carry out the replacement, but still it is better to replace the tangent with one letter - there is less risk of confusion.
Example 21
Find the indefinite integral
This is a do-it-yourself example.
Hold on, the championship rounds begin =)
Often in the integrand there is a "hodgepodge":
Example 22
Find the indefinite integral 
This integral initially contains a tangent, which immediately suggests an already familiar thought:
I will leave the artificial transformation at the very beginning and the rest of the steps without comment, since everything has already been said above.
A couple of creative examples for an independent solution:
Example 23
Find the indefinite integral 
Example 24
Find the indefinite integral 
Yes, in them, of course, you can lower the degrees of the sine, cosine, use the universal trigonometric substitution, but the solution will be much more efficient and shorter if it is drawn through tangents. Full solution and answers at the end of the lesson
An irrational function of a variable is a function that is formed from a variable and arbitrary constants using a finite number of operations of addition, subtraction, multiplication (raising to an integer power), division, and extraction of roots. An irrational function differs from a rational one in that an irrational function contains root extraction operations.
There are three main types of irrational functions whose indefinite integrals can be reduced to integrals of rational functions. These are integrals containing roots of arbitrary integer powers from a linear-fractional function (the roots can be of different degrees, but from the same linear-fractional function); integrals of the differential binomial and integrals with a square root of a square trinomial.
Important note. Roots are meaningful!
When calculating integrals containing roots, one often encounters expressions of the form , where is some function of the integration variable . In doing so, it should be borne in mind that . That is, for t > 0 , |t| = t. At t< 0 , |t| = - t . Therefore, when calculating such integrals, it is necessary to separately consider the cases t > 0 and t< 0 . This can be done by writing signs or where necessary. Assuming that the upper sign refers to the case t > 0 , and the lower one - to the case t< 0 . With further transformation, these signs, as a rule, are mutually reduced.
A second approach is also possible, in which the integrand and the result of integration can be considered as complex functions of complex variables. Then you can not follow the signs in the radical expressions. This approach is applicable if the integrand is analytic, that is, a differentiable function of a complex variable. In this case, both the integrand and its integral are multivalued functions. Therefore, after integration, when substituting numerical values, it is necessary to select a single-valued branch (Riemann surface) of the integrand, and for it select the corresponding branch of the integration result.
Fractional linear irrationality
These are integrals with roots of the same linear-fractional function:
,
where R is a rational function, are rational numbers, m 1 , n 1 , ..., m s , n s are integers, α, β, γ, δ are real numbers.
Such integrals are reduced to the integral of a rational function by substitution:
, where n is the common denominator of numbers r 1 , ..., r s .
The roots may not necessarily be from a linear-fractional function, but also from a linear one (γ = 0 , δ = 1), or from the integration variable x (α = 1 , β = 0 , γ = 0 , δ = 1).
Here are examples of such integrals:
,
.
Integrals from differential binomials
Integrals from differential binomials have the form:
,
where m, n, p are rational numbers, a, b are real numbers.
Such integrals reduce to integrals of rational functions in three cases.
1) If p is an integer. Substitution x = t N , where N is the common denominator of the fractions m and n .
2) If is an integer. Substitution a x n + b = t M , where M is the denominator of p .
3) If is an integer. Substitution a + b x - n = t M , where M is the denominator of p .
In other cases, such integrals are not expressed in terms of elementary functions.
Sometimes such integrals can be simplified using reduction formulas:
;
.
Integrals Containing the Square Root of a Square Trinomial
Such integrals have the form:
,
where R is a rational function. For each such integral, there are several methods for solving it.
1)
With the help of transformations lead to simpler integrals.
2)
Apply trigonometric or hyperbolic substitutions.
3)
Apply Euler substitutions.
Let's consider these methods in more detail.
1) Transformation of the integrand
Applying the formula and performing algebraic transformations, we bring the integrand to the form:
,
where φ(x), ω(x) are rational functions.
I type
Integral of the form:
,
where P n (x) is a polynomial of degree n.
Such integrals are found by the method of indefinite coefficients, using the identity:
.
Differentiating this equation and equating the left and right sides, we find the coefficients A i .
II type
Integral of the form:
,
where P m (x) is a polynomial of degree m.
Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the fraction should have an integer part.
III type
Here we make a substitution:
.
Then the integral will take the form:
.
Further, the constants α, β must be chosen such that the coefficients at t in the denominator vanish:
B = 0, B 1 = 0 .
Then the integral decomposes into the sum of integrals of two types:
,
,
which are integrated by substitutions:
u 2 \u003d A 1 t 2 + C 1,
v 2 \u003d A 1 + C 1 t -2.
2) Trigonometric and hyperbolic substitutions
For integrals of the form , a > 0
,
we have three main substitutions:
;
;
;
For integrals , a > 0
,
we have the following substitutions:
;
;
;
And, finally, for integrals , a > 0
,
substitutions are as follows:
;
;
;
3) Euler substitutions
Integrals can also be reduced to integrals of rational functions of one of the three Euler substitutions:
, for a > 0 ;
, for c > 0 ;
, where x 1 is the root of the equation a x 2 + b x + c = 0. If this equation has real roots.
Elliptic integrals
Finally, consider integrals of the form:
,
where R is a rational function, . Such integrals are called elliptic. In general, they are not expressed in terms of elementary functions. However, there are cases when there are relations between the coefficients A, B, C, D, E, in which such integrals are expressed in terms of elementary functions.
The following is an example related to recursive polynomials. The calculation of such integrals is performed using substitutions:
.
Example
Calculate integral:
.
Solution
We make a substitution.
.
Here, for x > 0
(u > 0
) we take the upper sign ′+ ′. For x< 0
(u< 0
) - lower ′- ′.
.
Answer
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, Lan, 2003.
In this section, we will consider a method for integrating rational functions. 7.1. Brief information about rational functions The simplest rational function is a polynomial of ti-th degree, i.e. a function of the form where are real constants, and a0 4 0. A polynomial Qn(x) whose coefficient a0 = 1 is called reduced. A real number b is called a root of the polynomial Qn(z) if Q„(b) = 0. It is known that each polynomial Qn(x) with real coefficients is uniquely decomposed into real factors of the form where p, q are real coefficients, and quadratic factors do not have real roots and, therefore, cannot be decomposed into real linear factors. Combining identical factors (if any) and assuming, for simplicity, the polynomial Qn(x) reduced, we can write down its factorization in the form where are natural numbers. Since the degree of the polynomial Qn(x) is equal to n, then the sum of all exponents a, /3, ..., A, added to the doubled sum of all exponents u, ..., q, is equal to n: The root a of the polynomial is called simple or single , if a = 1, and multiple if a > 1; the number a is called the multiplicity of the root a. The same applies to other polynomial roots. A rational function f(x) or a rational fraction is the ratio of two polynomials, and it is assumed that the polynomials Pm(x) and Qn(x) do not have common factors. A rational fraction is called proper if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. . If m p, then the rational fraction is called improper and in this case, dividing the numerator by the denominator according to the rule of dividing polynomials, it can be represented as where are some polynomials, and ^^ is a proper rational fraction. Example 1. A rational fraction is an improper fraction. Dividing by a "corner", we will have Hence. Here. and a proper fraction. Definition. The simplest (or elementary) fractions are called rational fractions of the following four types: where are real numbers, k is a natural number greater than or equal to 2, and the square trinomial x2 + px + q has no real roots, so -2 _2 is its discriminant In algebra the following theorem is proved. Theorem 3. A proper rational fraction with real coefficients whose denominator Qn(x) has the form is uniquely decomposed into a sum of simple fractions according to the rule Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions First Euler substitution Second Euler substitution Third Euler substitution In this expansion, some real constants, some of which may be equal to zero. To find these constants, the right side of equality (I) is reduced to a common denominator, and then the coefficients at the same powers of x in the numerators of the left and right sides are equated. This gives a system of linear equations from which the desired constants are found. . This method of finding unknown constants is called the method of indeterminate coefficients. Sometimes it is more convenient to apply another way of finding unknown constants, which consists in the fact that after equating the numerators, one obtains an identity for x, in which the argument x is given some values, for example, the values of the roots, resulting in equations for finding constants. It is especially convenient if the denominator Q„(x) has only real simple roots. Example 2. Decompose a rational fraction into simple fractions. This fraction is regular. We decompose the denominator into factors: Since the roots of the denominator are real and different, then, based on formula (1), the decomposition of the fraction into simplest ones will have the form Drive the right honor of “that equality to a common denominator and equating the numerators and its left and right parts, we get the identity or Unknown coefficient A. 2?, C we find in two ways. First way. Equating the coefficients at the same powers of x, t.v. with (free term), and the left and right parts of the identity, we get a linear system of equations for finding the unknown coefficients A, B, C: This system has a unique solution C The second way. Since the roots of the denominator are torn svv in i 0, we get 2 \u003d 2A, from where A * 1; g i 1, we get -1 * -B, from where 5 * 1; x i 2, we get 2 = 2C. whence C» 1, and the desired expansion has the form The denominator has two different dual roots: x\ \u003d 0 with a multiplicity of multiplicity 3. Therefore, the expansion of this non-simple fraction has the form Reducing the right side to a common denominator, we find or The first method. Equating the coefficients at the same powers of x in the left and right parts of the last identity. we obtain a linear system of equations. This system has a unique solution and the desired expansion will be the second method. In the resulting identity, setting x = 0, we obtain 1 a A2, or A2 = 1; field * gay x = -1, we get -3 i B), or Bj i -3. When substituting the found values of the coefficients A\ and B) and the identity will take the form or Putting x = 0, and then x = -I. we find that = 0, B2 = 0 and. so B\ = 0. Thus, we again get Example 4. Expand the rational fraction 4 into simple fractions The denominator of the fraction has no real roots, since the function x2 + 1 does not turn zero for any real values of x. Therefore, the expansion into simple fractions should have the form From here we get or. Equating the coefficients at Schinack powers of x in the left and right parts of the last equality, we will have whence we find and, therefore, It should be noted that in some cases expansions into simple fractions can be obtained faster and easier, acting in some other way, without using the method of indefinite coefficients. For example, to obtain the expansion of the fraction in example 3, you can add and subtract in the numerator 3x2 and perform division, as indicated below. 7.2. Integration of simple fractions As mentioned above, any improper rational fraction can be represented as the sum of some polynomial and a proper rational fraction (§7), and this representation is unique. Integration of a polynomial is not difficult, so consider the question of integrating a proper rational fraction. Since any proper rational fraction can be represented as a sum of simple fractions, its integration reduces to integrating simple fractions. Let us now consider the question of their integration. III. To find the integral of the simplest fraction of the third type, we select the full square of the binomial from the square trinomial: Since the second term, we set it equal to a2, where and then we make a substitution. Then, taking into account the linear properties of the integral, we find: Example 5. Find the integral 4 The integrand is the simplest fraction of the third type, since the square trinomial x1 + Ax + 6 has no real roots (its discriminant is negative: , and the numerator is a polynomial of the first degree. Therefore, we proceed as follows: 1) we select the full square in the denominator 2) we make a substitution (here 3) on * one integral To find the integral of the simplest fraction of the fourth type, we set, as above, . Then we get the Integral on the right side, denoted by A and transform it as follows: We integrate the integral on the right side by parts, setting whence or Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions First Euler substitution Second Euler substitution Third substitution Euler We have obtained the so-called recurrent formula, which allows us to find the integral Jk for any k = 2, 3,. .. . Indeed, the integral J\ is tabular: Assuming in the recursive formula, we find Knowing and assuming A = 3, we easily find Jj, and so on. In the final result, substituting everywhere instead of t and a their expressions in terms of x and the coefficients p and q, we obtain for the initial integral an expression for it in terms of x and the given numbers M, LG, p, q. Example 8. Find the integral in this means that the denominator has no real roots, and the numerator is a polynomial of the 1st degree. 1) We select a full square in the denominator 2) We make a substitution: The integral will take the form: Putting in the recurrent formula * = 2, a3 = 1. we will have, and, therefore, the desired integral is equal Returning to the variable x, we finally get 7.3. General case From the results of Secs. 1 and 2 of this section immediately follows an important theorem. Theorem! 4. An indefinite integral of any rational function always exists (on intervals in which the denominator of the fraction Q„(x) φ 0) and is expressed in terms of a finite number of elementary functions, namely, it is an algebraic sum. , rational fractions, natural logarithms and arctangents. So, to find an indefinite integral of a fractional-rational function, one should proceed in the following way: 1) if the rational fraction is incorrect, then the integer part is separated by dividing the numerator by the denominator, i.e., this function is represented as the sum of a polynomial and a proper rational fraction; 2) then the denominator of the obtained proper fraction is decomposed into the product of linear and quadratic factors; 3) this proper fraction decomposes into the sum of simple fractions; 4) using the linearity of the integral and the formula of item 2, we find the integrals of each term separately. Example 7. Find the integral M Since the denominator is a polynomial of the third degree, the integrand is an improper fraction. We single out the whole part in it: Therefore, we will have. The denominator of a regular fraction has phi different real roots: and therefore its decomposition into simple fractions has the form From here we find. Giving the argument x values equal to the roots of the denominator, we find from this identity that: Therefore, the desired integral will be equal to Example 8. Find the integral 4 1 of multiplicity 3, Therefore, the expansion of the integrand into simple fractions has the form Bringing the right side of this equality to a common denominator and reducing both sides of the equality by this denominator, we get or. We equate the coefficients at the same powers of x in the left and right parts of this identity: From here we find. Substituting the found values of the coefficients into the expansion, we will have Integrating, we find: Example 9. Find the integral 4 The denominator of the fraction has no real roots. Therefore, the expansion into simpler fractions of the integrand has the form From here or Equating the coefficients at the same powers of x in the left and right parts of this identity, we will have whence we find and, therefore, Remark. In the above example, the integrand can be represented as a sum of simple fractions in a simpler way, namely, in the numerator of the fraction, we select the bin in the denominator, and then perform a term-by-term division: §8. Integration of irrational functions A function of the form where Pm and u2 are polynomials of degrees type, respectively, in the variables u12,... is called a rational function in ubu2j... real constants, and Example 1, The function is a rational function of the variables r and y, since it represents both the ratio of a polynomial of the third degree and a polynomial of the fifth degree, and the yew function is not. In the case when the variables, in turn, are functions of the variable x: then the function ] is called a rational function of the functions of the Example. A function is a rational function of r and rvdkvlv Pryaivr 3. A function of the form is not a rational function of x and the radical y/r1 + 1, but it is a rational function of functions As the examples show, integrals of irrational functions are not always expressed in terms of elementary functions. For example, integrals often encountered in applications are not expressed in terms of elementary functions; these integrals are called elliptic integrals of the first and second kind, respectively. Let us consider those cases when the integration of irrational functions can be reduced with the help of some substitutions to the integration of rational functions. 1. Let it be required to find the integral where R(x, y) is a rational function of its arguments x and y; m £ 2 is a natural number; a, b, c, d are real constants satisfying the condition ad - bc ^ O (for ad - be = 0, the coefficients a and b are proportional to the coefficients c and d, and therefore the ratio does not depend on x; hence, in this case, the integrand function will be a rational function of the variable x, the integration of which was considered earlier). We make a change of variable in this integral by setting From here we express the variable x through a new variable We have x = - a rational function of t. Next, we find or, after simplification, Therefore, where L1 (t) is a rational function of *, since the rational funadium of a rational function, as well as the product of rational functions, are rational functions. We can integrate rational functions. Let Then the desired integral be equal to When. Go to the integral 4 The integrand* is a rational function of. Therefore, we put t = Then Integration of rational functions Brief information about rational functions Integration of simple fractions General case Integration of irrational functions First Euler substitution Second Euler substitution Third Euler substitution Thus, we get Primar 5. Find the integral the function can be represented as 1 _ 1_ whence it is clear that it is a rational function of: Considering this, we set. Therefore, 2. Consider intephs of the form where the subintegral function is such that replacing the radical \/ax2 + bx + c in it by y, we obtain the function R(x) y) - rational with respect to both arguments x and y. This integral is reduced to the integral of a rational function of another variable by Euler substitutions. 8.1. The first Euler substitution Let the coefficient a > 0. We set or From here we find x as a rational function of u, hence, Thus, the indicated substitution expresses rationally in terms of *. Therefore, we will have where Remark. The first Euler substitution can also be taken in the form Example 6. Find the integral we will find Therefore, we will have dx Euler substitution, show that Y 8.2. Euler's second substitution Let the trinomial ax2 + bx + c have different real roots R] and x2 (the coefficient can have any sign). In this case, we assume Since we get Since x, dxn y / ax2 + be + c are expressed rationally in terms of t, then the original integral reduces to the integral of a rational function, i.e. where Problem. Using the first Euler substitution, show that is a rational function of t. Example 7. Find the integral dx M function ] - x1 has different real roots. Therefore, we apply the second Euler substitution. From here we find Substituting the found expressions into Given? we get 8.3. The third substate of Euler Let the coefficient c > 0. We make a change of variable by setting. Note that to reduce the integral to the integral of a rational function, the first and second Euler substitutions suffice. Indeed, if the discriminant b2 -4ac > 0, then the roots of the square trinomial ax + bx + c are real, and in this case the second Euler substitution applies. If, then the sign of the trinomial ax2 + bx + c coincides with the sign of the coefficient a, and since the trinomial must be positive, then a > 0. In this case, the first Euler substitution applies. To find integrals of the form indicated above, it is not always expedient to use Euler substitutions, since other methods of integration can be found for them, leading to the goal more quickly. Let's consider some of these integrals. 1. To find integrals of the form, a right square is selected from the square of the th trinomial: where After that, a substitution is made and obtained where the coefficients a and P have different signs or they are both positive. When, as well as when a > 0, and the integral will be reduced to a logarithm, but if - to the arcsine. At. Find imtegrel 4 Since that. assuming, we get Prmmar 9. Find. I assumed x -, we will have 2. An integral of the form is reduced to the integral y from paragraph 1 as follows. Considering that the derivative ()" = 2, we select it in the numerator: th degree, can be found by the method of indeterminate coefficients, which consists in the following: Assume that the equality takes place Example 10. Mighty integral coefficients, we differentiate both sides of (1): Then we reduce the right side of equality (2) to a common denominator equal to the denominator of the left side, i.e. y/ax2 + bx + c, reducing both parts of (2) by which, we obtain an identity in both parts of which are polynomials of degree n. Equating the coefficients at the same powers of x in the left and right parts of (3), we obtain n + 1 equations, from which we find the required coefficients j4*(fc = 0,1,2,..., n Substituting their values into the right side of (1) and finding the integral + c we get the answer for this integral. Example 11. Find the integral We set Differentiating both suits of equality, we will have Bringing the right side to a common denominator and reducing both sides by it, we get the identity or. Equating the coefficients at the same powers of x, we arrive at a system of equations from which we find = Then we find the integral on the right side of equality (4): Therefore, the desired integral will be equal to
