The problem of finding the indefinite integral of a fractionally rational function comes down to integrating simple fractions. Therefore, we recommend that you first familiarize yourself with the section of the theory of decomposition of fractions into the simplest.

Example.

Solution.

Since the degree of the numerator of the integrand is equal to the degree of the denominator, we first select the whole part by dividing the polynomial by the polynomial with a column:

That's why, .

The decomposition of the resulting proper rational fraction into simpler fractions has the form . Hence,

The resulting integral is the integral of the simplest fraction of the third type. Looking ahead a little, we note that you can take it by subsuming it under the differential sign.

Because , That . That's why

Hence,

Now let's move on to describing methods for integrating simple fractions of each of the four types.

Integration of simple fractions of the first type

The direct integration method is ideal for solving this problem:

Example.

Solution.

Let's find the indefinite integral using the properties of the antiderivative, the table of antiderivatives and the integration rule.

Top of page

Integration of simple fractions of the second type

The direct integration method is also suitable for solving this problem:

Example.

Solution.

Top of page

Integration of simple fractions of the third type

First we present the indefinite integral as a sum:

We take the first integral by subsuming it under the differential sign:

That's why,

Let us transform the denominator of the resulting integral:

Hence,

The formula for integrating simple fractions of the third type takes the form:

Example.

Find the indefinite integral .

Solution.

We use the resulting formula:

If we didn’t have this formula, what would we do:

9. Integration of simple fractions of the fourth type

The first step is to put it under the differential sign:

The second step is to find an integral of the form . Integrals of this type are found using recurrence formulas. (See partitioning using recurrence formulas). The following recurrent formula is suitable for our case:

Example.

Find the indefinite integral

Solution.

For this type of integrand we use the substitution method. Let's introduce a new variable (see the section on integration of irrational functions):

After substitution we have:

We came to finding the integral of a fraction of the fourth type. In our case we have coefficients M = 0, p = 0, q = 1, N = 1 And n=3. We apply the recurrent formula:

After reverse replacement we get the result:

10. Integration of trigonometric functions.

Many problems come down to finding integrals of transcendental functions containing trigonometric functions. In this article we will group the most common types of integrands and use examples to consider methods for their integration.

Let's start by integrating sine, cosine, tangent and cotangent.

From the table of antiderivatives we immediately note that And .

The method of subsuming the differential sign allows you to calculate the indefinite integrals of the tangent and cotangent functions:

Top of page

Let's look at the first case, the second is absolutely similar.

Let's use the substitution method:

We came to the problem of integrating an irrational function. The substitution method will also help us here:

All that remains is to carry out the reverse replacement and t = sinx:

Top of page

You can learn more about the principles of finding them in the section integration using recurrent formulas. If you study the derivation of these formulas, you can easily take integrals of the form , Where m And n- integers.

Top of page

The most creativity comes when the integrand contains trigonometric functions with different arguments.

This is where the basic formulas of trigonometry come to the rescue. So write them down on a separate piece of paper and keep them before your eyes.

Example.

Find the set of antiderivatives of a function .

Solution.

The reduction formulas give And .

That's why

The denominator is the formula for the sine of the sum, therefore,

We arrive at the sum of three integrals.

Top of page

Integrands containing trigonometric functions can sometimes be reduced to fractional rational expressions using standard trigonometric substitution.

Let's write out trigonometric formulas expressing sine, cosine, tangent through the tangent of the half argument:

When integrating, we will also need the differential expression dx through the tangent of a half angle.

Because , That

That is, , where.

Example.

Find the indefinite integral .

Solution.

Let's use standard trigonometric substitution:

Thus, .

Decomposing the integrand into simple fractions leads us to the sum of two integrals:

All that remains is to carry out the reverse replacement:

11. Recurrence formulas are formulas that express n The th member of the sequence through the previous members. They are often used when finding integrals.

We do not aim to list all recurrence formulas, but want to give the principle of their derivation. The derivation of these formulas is based on the transformation of the integrand and the application of the method of integration by parts.

For example, the indefinite integral can be taken using the recurrence formula .

Derivation of the formula:

Using trigonometry formulas, we can write:

We find the resulting integral using the method of integration by parts. As a function u(x) let's take cosx, hence, .

That's why,

We return to the original integral:

That is,

That's what needed to be shown.

The following recurrence formulas are derived similarly:

Example.

Find the indefinite integral.

Solution.

We use the recurrent formula from the fourth paragraph (in our example n=3):

Since from the table of antiderivatives we have , That

All of the above in the previous paragraphs allows us to formulate the basic rules for the integration of rational fractions.

1. If a rational fraction is improper, then it is represented as the sum of a polynomial and a proper rational fraction (see paragraph 2).

This reduces the integration of an improper rational fraction to the integration of a polynomial and a proper rational fraction.

2. Factor the denominator of the proper fraction.

3. A proper rational fraction is decomposed into the sum of simple fractions. This reduces the integration of a proper rational fraction to the integration of simple fractions.

Let's look at examples.

Example 1. Find .

Solution. Below the integral is an improper rational fraction. Selecting the whole part, we get

Hence,

Noting that , let us expand the proper rational fraction

to simple fractions:

(see formula (18)). That's why

Thus, we finally have

Example 2. Find

Solution. Below the integral is a proper rational fraction.

Expanding it into simple fractions (see formula (16)), we obtain

Integration of a fractional-rational function.
Uncertain coefficient method

We continue to work on integrating fractions. We have already looked at integrals of some types of fractions in the lesson, and this lesson, in a sense, can be considered a continuation. To successfully understand the material, basic integration skills are required, so if you have just started studying integrals, that is, you are a beginner, then you need to start with the article Indefinite integral. Examples of solutions.

Oddly enough, now we will be engaged not so much in finding integrals, but... in solving systems of linear equations. In this regard urgently I recommend attending the lesson. Namely, you need to be well versed in substitution methods (“the school” method and the method of term-by-term addition (subtraction) of system equations).

What is a fractional rational function? In simple words, a fractional-rational function is a fraction whose numerator and denominator contain polynomials or products of polynomials. Moreover, the fractions are more sophisticated than those discussed in the article Integrating Some Fractions.

Integrating a Proper Fractional-Rational Function

Immediately an example and a typical algorithm for solving the integral of a fractional-rational function.

Example 1

Step 1. The first thing we ALWAYS do when solving an integral of a fractional rational function is to clarify the following question: is the fraction proper? This step is performed verbally, and now I will explain how:

First we look at the numerator and find out senior degree polynomial:

The leading power of the numerator is two.

Now we look at the denominator and find out senior degree denominator. The obvious way is to open the brackets and bring similar terms, but you can do it simpler, in each find the highest degree in brackets

and mentally multiply: - thus, the highest degree of the denominator is equal to three. It is quite obvious that if we actually open the brackets, we will not get a degree greater than three.

Conclusion: Major degree of numerator STRICTLY is less than the highest power of the denominator, which means the fraction is proper.

If in this example the numerator contained the polynomial 3, 4, 5, etc. degrees, then the fraction would be wrong.

Now we will consider only the correct fractional rational functions. The case when the degree of the numerator is greater than or equal to the degree of the denominator will be discussed at the end of the lesson.

Step 2. Let's factorize the denominator. Let's look at our denominator:

Generally speaking, this is already a product of factors, but, nevertheless, we ask ourselves: is it possible to expand something else? The object of torture will undoubtedly be the square trinomial. Solving the quadratic equation:

The discriminant is greater than zero, which means that the trinomial really can be factorized:

General rule: EVERYTHING in the denominator CAN be factored - factored

Let's begin to formulate a solution:

Step 3. Using the method of indefinite coefficients, we expand the integrand into a sum of simple (elementary) fractions. Now it will be clearer.

Let's look at our integrand function:

And, you know, somehow an intuitive thought pops up that it would be nice to turn our large fraction into several small ones. For example, like this:

The question arises, is it even possible to do this? Let us breathe a sigh of relief, the corresponding theorem of mathematical analysis states – IT IS POSSIBLE. Such a decomposition exists and is unique.

There's just one catch, the odds are Bye We don’t know, hence the name – the method of indefinite coefficients.

As you guessed, subsequent body movements are like that, don’t cackle! will be aimed at just RECOGNIZING them - to find out what they are equal to.

Be careful, I will explain in detail only once!

So, let's start dancing from:

On the left side we reduce the expression to a common denominator:

Now we can safely get rid of the denominators (since they are the same):

On the left side we open the brackets, but do not touch the unknown coefficients for now:

At the same time, we repeat the school rule for multiplying polynomials. When I was a teacher, I learned to pronounce this rule with a straight face: In order to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial.

From the point of view of a clear explanation, it is better to put the coefficients in brackets (although I personally never do this in order to save time):

We compose a system of linear equations.
First we look for senior degrees:

And we write the corresponding coefficients into the first equation of the system:

Remember the following point well. What would happen if there were no s on the right side at all? Let's say, would it just show off without any square? In this case, in the equation of the system it would be necessary to put a zero on the right: . Why zero? But because on the right side you can always assign this same square with zero: If on the right side there are no variables and/or a free term, then we put zeros on the right sides of the corresponding equations of the system.

We write the corresponding coefficients into the second equation of the system:

And finally, mineral water, we select free members.

Eh...I was kind of joking. Jokes aside - mathematics is a serious science. In our institute group, no one laughed when the assistant professor said that she would scatter the terms along the number line and choose the largest ones. Let's get serious. Although... whoever lives to see the end of this lesson will still smile quietly.

The system is ready:

We solve the system:

(1) From the first equation we express and substitute it into the 2nd and 3rd equations of the system. In fact, it was possible to express (or another letter) from another equation, but in this case it is advantageous to express it from the 1st equation, since there the smallest odds.

(2) We present similar terms in the 2nd and 3rd equations.

(3) We add the 2nd and 3rd equations term by term, obtaining the equality , from which it follows that

(4) We substitute into the second (or third) equation, from where we find that

(5) Substitute and into the first equation, obtaining .

If you have any difficulties with the methods of solving the system, practice them in class. How to solve a system of linear equations?

After solving the system, it is always useful to check - substitute the found values every equation of the system, as a result everything should “converge”.

Almost there. The coefficients were found, and:

The finished job should look something like this:

As you can see, the main difficulty of the task was to compose (correctly!) and solve (correctly!) a system of linear equations. And at the final stage, everything is not so difficult: we use the linearity properties of the indefinite integral and integrate. Please note that under each of the three integrals we have a “free” complex function; I talked about the features of its integration in the lesson Variable change method in indefinite integral.

Check: Differentiate the answer:

The original integrand function has been obtained, which means that the integral has been found correctly.
During the verification, we had to reduce the expression to a common denominator, and this is not accidental. The method of indefinite coefficients and reducing an expression to a common denominator are mutually inverse actions.

Example 2

Find the indefinite integral.

Let's return to the fraction from the first example: . It is easy to notice that in the denominator all the factors are DIFFERENT. The question arises, what to do if, for example, the following fraction is given: ? Here we have degrees in the denominator, or, mathematically, multiples. In addition, there is a quadratic trinomial that cannot be factorized (it is easy to verify that the discriminant of the equation is negative, so the trinomial cannot be factorized). What to do? The expansion into a sum of elementary fractions will look something like with unknown coefficients at the top or something else?

Example 3

Introduce a function

Step 1. Checking if we have a proper fraction
Major numerator: 2
Highest degree of denominator: 8
, which means the fraction is correct.

Step 2. Is it possible to factor something in the denominator? Obviously not, everything is already laid out. The square trinomial cannot be expanded into a product for the reasons stated above. Hood. Less work.

Step 3. Let's imagine a fractional-rational function as a sum of elementary fractions.
In this case, the expansion has the following form:

Let's look at our denominator:
When decomposing a fractional-rational function into a sum of elementary fractions, three fundamental points can be distinguished:

1) If the denominator contains a “lonely” factor to the first power (in our case), then we put an indefinite coefficient at the top (in our case). Examples No. 1, 2 consisted only of such “lonely” factors.

2) If the denominator has multiple multiplier, then you need to decompose it like this:
- that is, sequentially go through all the degrees of “X” from the first to the nth degree. In our example there are two multiple factors: and , take another look at the expansion I gave and make sure that they are expanded exactly according to this rule.

3) If the denominator contains an indecomposable polynomial of the second degree (in our case), then when decomposing in the numerator you need to write a linear function with undetermined coefficients (in our case with undetermined coefficients and ).

In fact, there is another 4th case, but I will keep silent about it, since in practice it is extremely rare.

Example 4

Introduce a function as a sum of elementary fractions with unknown coefficients.

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.
Follow the algorithm strictly!

If you understand the principles by which you need to expand a fractional-rational function into a sum, you can chew through almost any integral of the type under consideration.

Example 5

Find the indefinite integral.

Step 1. Obviously the fraction is correct:

Step 2. Is it possible to factor something in the denominator? Can. Here is the sum of cubes . Factor the denominator using the abbreviated multiplication formula

Step 3. Using the method of indefinite coefficients, we expand the integrand into a sum of elementary fractions:

Please note that the polynomial cannot be factorized (check that the discriminant is negative), so at the top we put a linear function with unknown coefficients, and not just one letter.

We bring the fraction to a common denominator:

Let's compose and solve the system:

(1) We express from the first equation and substitute it into the second equation of the system (this is the most rational way).

(2) We present similar terms in the second equation.

(3) We add the second and third equations of the system term by term.

All further calculations are, in principle, oral, since the system is simple.

(1) We write down the sum of fractions in accordance with the found coefficients.

(2) We use the linearity properties of the indefinite integral. What happened in the second integral? You can familiarize yourself with this method in the last paragraph of the lesson. Integrating Some Fractions.

(3) Once again we use the properties of linearity. In the third integral we begin to isolate the complete square (penultimate paragraph of the lesson Integrating Some Fractions).

(4) We take the second integral, in the third we select the complete square.

(5) Take the third integral. Ready.

Examples of integrating rational functions (fractions) with detailed solutions are considered.

Content

Example 1

Calculate the integral:
.

Here, under the integral sign there is a rational function, since the integrand is a fraction of polynomials. Denominator polynomial degree ( 3 ) is less than the degree of the numerator polynomial ( 4 ). Therefore, first you need to select the whole part of the fraction.

1. Let's select the whole part of the fraction. Divide x 4 by x 3 - 6 x 2 + 11 x - 6:

From here
.

2. Let's factorize the denominator of the fraction. To do this, you need to solve the cubic equation:
.
6
1, 2, 3, 6, -1, -2, -3, -6 .
Let's substitute x = 1 :
.

1 . Divide by x - 1 :

From here
.
Solving a quadratic equation.
.
The roots of the equation are: , .
Then
.

3. Let's break down the fraction into its simplest form.

.

So we found:
.
Let's integrate.

Example 2

Calculate the integral:
.

Here the numerator of the fraction is a polynomial of degree zero ( 1 = x 0). The denominator is a polynomial of the third degree. Because the 0 < 3 , then the fraction is correct. Let's break it down into simple fractions.

1. Let's factorize the denominator of the fraction. To do this, you need to solve the third degree equation:
.
Let's assume that it has at least one whole root. Then it is a divisor of the number 3 (member without x). That is, the whole root can be one of the numbers:
1, 3, -1, -3 .
Let's substitute x = 1 :
.

So, we have found one root x = 1 . Divide x 3 + 2 x - 3 on x - 1 :

So,
.

Solving the quadratic equation:
x 2 + x + 3 = 0.
Find the discriminant: D = 1 2 - 4 3 = -11. Since D< 0 , then the equation has no real roots. Thus, we obtained the factorization of the denominator:
.

2.
.
(x - 1)(x 2 + x + 3):
(2.1) .
Let's substitute x = 1 . Then x - 1 = 0 ,
.

Let's substitute in (2.1) x = 0 :
1 = 3 A - C;
.

Let's equate to (2.1) coefficients for x 2 :
;
0 = A + B;
.

3. Let's integrate.
(2.2) .
To calculate the second integral, we select the derivative of the denominator in the numerator and reduce the denominator to the sum of squares.

;
;
.

Calculate I 2 .

.
Since the equation x 2 + x + 3 = 0 has no real roots, then x 2 + x + 3 > 0. Therefore, the modulus sign can be omitted.

We deliver to (2.2) :
.

Example 3

Calculate the integral:
.

Here under the integral sign there is a fraction of polynomials. Therefore, the integrand is a rational function. The degree of the polynomial in the numerator is equal to 3 . The degree of the polynomial of the denominator of the fraction is equal to 4 . Because the 3 < 4 , then the fraction is correct. Therefore, it can be decomposed into simple fractions. But to do this you need to factorize the denominator.

1. Let's factorize the denominator of the fraction. To do this, you need to solve the fourth degree equation:
.
Let's assume that it has at least one whole root. Then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's substitute x = -1 :
.

So, we have found one root x = -1 . Divide by x - (-1) = x + 1:

So,
.

Now we need to solve the third degree equation:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2 (member without x). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Let's substitute x = -1 :
.

So, we found another root x = -1 . It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.

Since the equation x 2 + 2 = 0 has no real roots, then we get the factorization of the denominator:
.

2. Let's break down the fraction into its simplest form. We are looking for an expansion in the form:
.
We get rid of the denominator of the fraction, multiply by (x + 1) 2 (x 2 + 2):
(3.1) .
Let's substitute x = -1 . Then x + 1 = 0 ,
.

Let's differentiate (3.1) :

;

.
Let's substitute x = -1 and take into account that x + 1 = 0 :
;
; .

Let's substitute in (3.1) x = 0 :
0 = 2 A + 2 B + D;
.

Let's equate to (3.1) coefficients for x 3 :
;
1 = B + C;
.

So, we have found the decomposition into simple fractions:
.

3. Let's integrate.

.

Examples of solutions

Example 1
Find the integral of the rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$
Solution
The fraction is proper and the polynomial has only complex roots. Therefore, we select a complete square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We fold a complete square and place it under the differential sign $ x-5 $: $$ = \int \frac(dx)((x-5)^2 - 9) = \int \frac(d(x-5))((x-5)^2-9) = $$ Using the table of integrals we obtain: $$ = \frac(1)(2 \cdot 3) \ln \bigg \| \frac(x-5 - 3)(x-5 + 3) \bigg \| + C = \frac(1)(6) \ln \bigg \|\frac(x-8)(x-2) \bigg \| +C$$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner!
Answer
$$ \int \frac(dx)(x^2-10x+16) = \frac(1)(6) \ln \bigg \|\frac(x-8)(x-2) \bigg \| +C$$

Example 2
Perform integration of rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$
Solution
Let's solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ We write down the roots: $$ x_1 = \frac(-5-7)(2) = -6; x_2 = \frac(-5+7)(2) = 1 $$ Taking into account the obtained roots, we transform the integral: $$ \int \frac(x+2)(x^2+5x-6) dx = \int \frac(x+2)((x-1)(x+6)) dx = $$ We perform the expansion of a rational fraction: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ We equate the numerators and find the coefficients $ A $ and $ B $: $$ A(x+6)+B(x-1)=x+2 $$ $$ Ax + 6A + Bx - B = x + 2 $$ $$ \begin(cases) A + B = 1 \\ 6A - B = 2 \end(cases) $$ $$ \begin(cases) A = \frac(3)(7) \\ B = \frac(4)(7) \end(cases) $$ We substitute the found coefficients into the integral and solve it: $$ \int \frac(x+2)((x-1)(x+6))dx = \int \frac(\frac(3)(7))(x-1) dx + \int \frac (\frac(4)(7))(x+6) dx = $$ $$ = \frac(3)(7) \int \frac(dx)(x-1) + \frac(4)(7) \int \frac(dx)(x+6) = \frac(3) (7) \ln \|x-1\| + \frac(4)(7) \ln \|x+6\| +C$$
Answer
$$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln \|x-1\| + \frac(4)(7) \ln \|x+6\| +C$$