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

Content

See also: The roots of a quadratic equation

Here we provide detailed solutions to three examples of integrating the following rational fractions:
, , .

Example 1

Calculate integral:
.

Here, under the integral sign there is a rational function, since the integrand is a fraction of polynomials. The degree of the denominator polynomial ( 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 take the integer part of the fraction. Divide x 4 on x 3 - 6 x 2 + 11 x - 6:

From here
.

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

1 . Divide by x - 1 :

From here
.
We decide quadratic equation.
.
Equation roots: , .
Then
.

3. Let's decompose the fraction into simple ones.

.

So we found:
.
Let's integrate.

Example 2

Calculate integral:
.

Here in the numerator of the fraction is a polynomial of degree zero ( 1 = x0). The denominator is a third degree polynomial. Because the 0 < 3 , then the fraction is correct. Let's break it down into simple fractions.

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

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

So,
.

We solve the quadratic equation:
x 2 + x + 3 = 0.
Find the discriminant: D = 1 2 - 4 3 = -11. Because D< 0 , then the equation has no real roots. Thus, we have obtained the decomposition of the denominator into factors:
.

2.
.
(x - 1)(x 2 + x + 3):
(2.1) .
Substitute x = 1 . Then x- 1 = 0 ,
.

Substitute in (2.1) x= 0 :
1 = 3 A - C;
.

Equate in (2.1) coefficients at 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 module sign can be omitted.

We deliver to (2.2) :
.

Example 3

Calculate integral:
.

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

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

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

So,
.

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

So, we have 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 decompose the fraction into simple ones. We are looking for a decomposition in the form:
.
We get rid of the denominator of the fraction, multiply by (x + 1) 2 (x 2 + 2):
(3.1) .
Substitute x = -1 . Then x + 1 = 0 ,
.

Differentiate (3.1) :

;

.
Substitute x = -1 and take into account that x + 1 = 0 :
;
; .

Substitute in (3.1) x= 0 :
0 = 2A + 2B + D;
.

Equate in (3.1) coefficients at x 3 :
;
1=B+C;
.

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

3. Let's integrate.

.

Integration of a fractional-rational function.
Method of undetermined coefficients

We continue to work on integrating fractions. We have already considered integrals of some types of fractions in the lesson, and in a sense this lesson 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 teapot, then you need to start with the article Indefinite integral. Solution examples.

Oddly enough, now we will be engaged not so much in finding integrals as in ... solving systems linear equations. In this connection strongly I recommend visiting the lesson Namely, you need to be well versed in the 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 in the numerator and denominator of which are polynomials or products of polynomials. At the same time, fractions are more sophisticated than those discussed in the article. Integration of some fractions.

Integration of the correct 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 rational-fractional function is find out next question: is the fraction correct? This step is done orally, and now I will explain how:

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

The highest power of the numerator is two.

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

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

Conclusion: Highest power of the numerator STRICTLY less than the highest power of the denominator, then the fraction is correct.

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

Now we will consider only proper fractional-rational functions. The case when the degree of the numerator is greater than or equal to the degree of the denominator, we will analyze at the end of the lesson.

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

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

The discriminant is greater than zero, which means that the trinomial is indeed factorized:

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

Let's start making a decision:

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

Let's look at our integrand function:

And, you know, an intuitive thought somehow slips through 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's breathe a sigh of relief, the corresponding theorem of mathematical analysis states - IT IS POSSIBLE. Such a decomposition exists and is unique.

There is only one catch, the coefficients we Bye we do not know, hence the name - the method of indefinite coefficients.

You guessed it, the subsequent gestures so, do not cackle! will be aimed at just LEARNING them - to find out what they are equal to.

Be careful, I explain in detail once!

So, let's start dancing from:

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

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

On the left side, we open the brackets, while we do not touch the unknown coefficients yet:

At the same time we repeat school rule multiplication of polynomials. When I was a teacher, I learned to say 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 in the first equation of the system:

Well remember the following nuance. What would happen if the right side did not exist at all? Say, would it just show off without any square? In this case, in the equation of the system, it would be necessary to put zero on the right: . Why zero? And because on the right side you can always attribute this same square with zero: If there are no variables or (and) a free term on the right side, then we put zeros on the right sides of the corresponding equations of the system.

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

And, finally, mineral water, we select free members.

Eh, ... I was 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 members along a number line and choose the largest of them. Let's get serious. Although ... whoever lives to see the end of this lesson will still smile quietly.

System 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, while obtaining the equality , from which it follows that

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

(5) We substitute and into the first equation, getting .

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

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

Almost arrived. The coefficients are found, while:

A clean 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 properties of linearity indefinite integral and integrate. I draw your attention to the fact 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 was obtained, which means that the integral was found correctly.
During the verification, it was necessary to bring the expression to a common denominator, and this is not accidental. The method of indefinite coefficients and bringing the expression to a common denominator are mutually inverse actions.

Example 2

Find the indefinite integral.

Let's go back to the fraction from the first example: . It is easy to see that in the denominator all factors are DIFFERENT. The question arises, what to do if, for example, such a fraction is given: ? Here we have degrees in the denominator, or, in mathematical terms, multiple factors. In addition, there is an indecomposable square trinomial (it is easy to verify that the discriminant of the equation is negative, so the trinomial cannot be factored in any way). What to do? The expansion into a sum of elementary fractions will look like with unknown coefficients at the top or some other way?

Example 3

Submit a function

Step 1. Checking if we have a correct fraction
Highest power of the numerator: 2
Highest denominator: 8
, so the fraction is correct.

Step 2 Can anything be factored in the denominator? Obviously not, everything is already laid out. Square trinomial does not expand into a product for the reasons indicated above. Good. Less work.

Step 3 Let us represent a fractional-rational function as a sum of elementary fractions.
In this case, the decomposition 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 in the first degree (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 contains multiple multiplier, then you need to decompose as follows:
- that is, sequentially sort 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 decomposition I have given and make sure that they are decomposed exactly according to this rule.

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

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

Example 4

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

This is an example for independent solution. Complete Solution and the answer at the end of the lesson.
Strictly follow the algorithm!

If you have figured out the principles by which you need to decompose a fractional-rational function into a sum, then you can crack almost any integral of the type under consideration.

Example 5

Find the indefinite integral.

Step 1. Obviously, the fraction is correct:

Step 2 Can anything be factored in the denominator? Can. Here is the sum of cubes . Factoring 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:

Note that the polynomial is indecomposable (check that the discriminant is negative), so at the top we put a linear function with unknown coefficients, and not just a single letter.

We bring the fraction to a common denominator:

Let's create and solve the system:

(1) From the first equation, we express and substitute 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, in principle, are 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 find this method in the last paragraph of the lesson. Integration of some fractions.

(3) Once again we use the properties of linearity. In the third integral, we begin to isolate full square(penultimate paragraph of the lesson Integration of some fractions).

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

(5) We take the third integral. Ready.

The fraction is called correct if the highest power of the numerator is less than the highest power of the denominator. The integral of a proper rational fraction has the form:

$$ \int \frac(mx+n)(ax^2+bx+c)dx $$

The formula for integrating rational fractions depends on the roots of the polynomial in the denominator. If the polynomial $ ax^2+bx+c $ has:

Only complex roots, then it is necessary to select a full square from it: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(mx+n)(x^2 \pm a ^2) $$
Different real roots $ x_1 $ and $ x_2 $, then you need to expand the integral and find the indefinite coefficients $ A $ and $ B $: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)(x-x_1) dx + \int \frac(B)(x-x_2) dx $$
One multiple root $ x_1 $, then we expand the integral and find the indefinite coefficients $ A $ and $ B $ for this formula: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)((x-x_1)^2)dx + \int \frac(B)(x-x_1) dx $$

If the fraction is wrong, that is, the highest degree in the numerator is greater than or equal to the highest degree of the denominator, then first it must be reduced to correct mind by dividing the polynomial from the numerator by the polynomial from the denominator. In this case, the formula for integrating a rational fraction is:

$$ \int \frac(P(x))(ax^2+bx+c)dx = \int Q(x) dx + \int \frac(mx+n)(ax^2+bx+c)dx $$

Solution examples

Example 1
Find the integral of a rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$
Solution
The fraction is regular and the polynomial has only complex roots. Therefore, we select a full square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We collapse the full square and sum 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 get: $$ = \frac(1)(2 \cdot 3) \ln \bigg \| \frac(x-5 - 3)(x-5 + 3) \bigg \| + C = \frac(1)(6) \ln \bigg \|\frac(x-8)(x-2) \bigg \| +C$$ If you cannot solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the progress of the calculation and gather information. This will help you get a credit from the 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
Integrate rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$
Solution
Solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ Let's 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: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ 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
$$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln \|x-1\| + \frac(4)(7) \ln \|x+6\| +C$$

Example 1

Example 2

Example 3

Integration of a fractional-rational function. Method of undetermined coefficients

Integration of the correct fractional-rational function

Solution examples

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Integration of a fractional-rational function.
Method of undetermined coefficients