Integrating the simplest (elementary) fractions. Integrating rational functions and method of indefinite coefficients

"Mathematics just as an artist or poet creates patterns. And if his patterns are more stable, only because they are composed of ideas ... patterns of mathematics just like patterns of an artist or poet, should be beautiful; Ideas just like colors or words, must match each other. Beauty is the first requirement: there is no place for ugly mathematics in the world».

G.H.hardi

In the first chapter, it was noted that there are first-shaped pretty simple functions that can already be expressed through elementary functions. In this regard, the classes of functions that can be exactly said to say that their primary - elementary functions acquire tremendous practical importance. To such a class of functions rational functionsrepresenting the ratio of two algebraic polynomials. Many tasks lead to the integration of rational fractions. Therefore, it is very important to be able to integrate such functions.

2.1.1. Fractional rational functions

Rational fraction (or fractional rational function) The attitude of two algebraic polynomials is called:

where and are polynomials.

Recall that polynomial (polynomial, whole rational function) n.degree called the type of type

where - Actual numbers. For example,

- the first degree polynomial;

- a polynomial of the fourth degree, etc.

Rational fraction (2.1.1) called rightif the degree is lower than the degree, i.e. n.<m.otherwise the fraction is called wrong.

Any incorrect fraction can be represented as a number of polynomials (integer part) and proper fraction (fractional part). The allocation of whole and fractional parts of the wrong fraction can be made according to the rules of dividing polynomials "corner".

Example 2.1.1. Select the whole and fractional parts of the following incorrect rational fractions:

but) b) .

Decision . a) using the division algorithm "Corner", we get

Thus, we get

.

b) Here we also use the division algorithm "Corner":

As a result, we get

.

Let's summarize. The indefinite integral from the rational fraction in the general case can be submitted to the amount of integrals from the polynomial and from the correct rational fraction. Finding primary from polynomials does not represent difficulties. Therefore, in the future we will consider mainly the right rational fractions.

2.1.2. The simplest rational fractions and their integration

Among the correct rational fractions, four types are distinguished that belong to simplest (elementary) rational fractions:

3) ,

4) ,

where is an integer . Square Treechlen It does not have valid roots.

The integration of the simplest fractions of the 1st and 2nd type does not represent large difficulties:

, (2.1.3)

. (2.1.4)

We will now consider the integration of the simplest fractions of the 3rd type, and we will not consider the fraction of the 4th type.

Let's start with the integrals of the form

.

This integral is typically calculated by highlighting a complete square in the denominator. The result is a table integral of the following type

or .

Example 2.1.2. Find integrals:

but) b) .

Decision . a) We highlight the full square from the square triple:

From here to find

b) highlighting a full square from a square triple, we get:

In this way,

.

To find the integral

it can be isolated in the numerator of the denominator derivative and decompose the integral for the amount of two integrals: the first substitution of them comes down to the form

,

and the second to the above.

Example 2.1.3. Find integrals:

.

Decision . notice, that . We highlight the denominator derivative in the numerator:

The first integral is calculated by substitution :

In the second integral, we highlight a full square in the denominator

Finally, get

2.1.3. Decomposition of the correct rational fraction
on the sum of the simplest fractions

Any correct rational fraction It can be submitted the only way as the sum of the simplest fractions. For this, the denominator needs to be decomposed on multipliers. From the highest algebra it is known that each polynomial with valid coefficients

As is known, any rational function from some variable x can be decomposed on polynomials and the simplest, elementary, fractions. There are four types of simple frains:
1) ;
2) ;
3) ;
4) .
Here a, a, b, b, c is valid numbers. Equation X. 2 + BX + C \u003d 0 It does not have valid roots.

Integrating the fractions of the first two types

The integration of the first two fractions is performed using the following formulas from the integral table:
,
, N ≠ - 1 .

1. Integration of the finish of the first type

The fraction of the first type of substitution T \u003d X - A is provided to the tabular integral:
.

2. Integration of the fraction of the second type

The fraction of the second type is given to the tabular integral of the same substitution T \u003d X - A:

.

3. Integration of the fraction of the third type

Consider the integral of the fraction of the third type:
.
We will calculate it in two receptions.

3.1. Step 1. Highlight a denominator derivative in the numerator

We highlight the fraction derivative from the denominator in the numerator. Denote: u \u003d x 2 + BX + C. Differentiating: u '\u003d 2 x + b. Then
;
.
But
.
We lowered the module sign, because.

Then:
,
Where
.

3.2. Step 2. Calculate the integral with a \u003d 0, b \u003d 1

Now we calculate the remaining integral:
.

We give a denomote driver to the sum of the squares:
,
where.
We believe that equation x 2 + BX + C \u003d 0 It does not have roots. Therefore .

Make a substitution
,
.
.

So,
.

Thus, we found an integral from the fraction of the third type:

,
where.

4. Integration of fourth-type fraction

And finally, consider the integral from the fourth fraction:
.
Calculate it in three receptions.

4.1) We allocate a denominator derivative in the numerator:
.

4.2) Calculate the integral
.

4.3) Calculate the integrals
,
Using the brief formula:
.

4.1. Step 1. Selection in the numerator of the denominator

We highlight a derivative of the denominator in the numerator, as we did in. Denote u \u003d x 2 + BX + C. Differentiating: u '\u003d 2 x + b. Then
.

.
But
.

Finally we have:
.

4.2. Step 2. Calculation of the integral with n \u003d 1

Calculate the integral
.
Its calculation is set out in.

4.3. Step 3. Output of the formula of bringing

Now consider the integral
.

We give a square triple to the sum of the squares:
.
Here .
Make a substitution.
.
.

We carry out transformations and integrate in parts.




.

Multiply by 2 (n - 1):
.
Return to x and i n.
,
;
;
.

So, for I N, we received a formula of bringing:
.
Sequentially applying this formula, we will reduce the integral I n to I 1 .

Example

Calculate integral

Decision

1. We highlight the denominator derivative in the numerator.
;
;


.
Here
.

2. Calculate the integral from the simplest fraction.

.

3. We use the formula of bringing:

For integral.
In our case b \u003d 1 , C \u003d. 1 , 4 C - B 2 \u003d 3. We write this formula for n \u003d 2 and n \u003d 3 :
;
.
From here

.

Finally we have:

.
We find the coefficient at.
.

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

Example 1.

Calculate the integral:
.

Decision

Here, under the integral sign there is a rational function, since the integrand is a fraction of polynomials. The degree of polynomial of the denominator ( 3 ) less than the degree of polynomial numerator ( 4 ). Therefore, at first it is necessary to allocate the whole part of the fraction.

1. We highlight the whole part of the fraci. Delim X. 4 on X. 3 - 6 x 2 + 11 x - 6:

From here
.

2. Spatulate the denomoter of the fractions on multipliers. To do this, solve a cubic equation:
.
6
1, 2, 3, 6, -1, -2, -3, -6 .
Substitute x \u003d 1 :
.

1 . We divide on X - 1 :

From here
.
We solve the square equation.
.
Root equations: ,.
Then
.

3. We decompose the fraction on the simplest.

.

So, we found:
.
We integrate.

Answer

Example 2.

Calculate the integral:
.

Decision

Here in the numerator of the fraction - a polynomial of zero degree ( 1 \u003d x 0). In the denominator - a polynomial of the third degree. Insofar as 0 < 3 , the crushing is correct. Spread it on the simplest fraction.

1. Spatulate the denomoter of the fractions on multipliers. To do this, it is necessary to solve the equation of the third degree:
.
Suppose it has at least one whole root. Then he is a divider of the number 3 (Member without X). That is, the whole root can be one of the numbers:
1, 3, -1, -3 .
Substitute x \u003d 1 :
.

So, we found one root X \u003d 1 . Delim X. 3 + 2 x - 3 on x - 1 :

So,
.

We solve the square equation:
x. 2 + x + 3 \u003d 0.
We find discriminant: D \u003d 1 2 - 4 · 3 \u003d -11. Since D.< 0 The equation does not have valid roots. Thus, we received a decomposition of the denominator for multipliers:
.

2.
.
(x - 1) (x 2 + x + 3):
(2.1) .
Substitute x \u003d 1 . Then X - 1 = 0 ,
.

Substitute B. (2.1) x \u003d. 0 :
1 \u003d 3 A - C;
.

Ensure B. (2.1) Coefficients at X. 2 :
;
0 \u003d a + b;
.

.

3. We integrate.
(2.2) .
To calculate the second integral, select the denominator derivative in the numerator and give the denominator to the sum of the squares.

;
;
.

Calculate I. 2 .


.
Since equation X. 2 + x + 3 \u003d 0 does not have valid roots, then x 2 + x + 3\u003e 0. Therefore, the module sign can be omitted.

Supply in (2.2) :
.

Answer

Example 3.

Calculate the integral:
.

Decision

Here, under the sign of the integral, it is worth a fraction from polynomials. Therefore, the integrand is a rational function. The degree of polynomial in the numerator is equal 3 . The degree of polynomial of the denominator of the fraction is equal 4 . Insofar as 3 < 4 , the crushing is correct. Therefore, it can be laid on the simplest fraction. But for this you need to decompose the denominator for multipliers.

1. Spatulate the denomoter of the fractions on multipliers. To do this, it is necessary to solve the fourth degree equation:
.
Suppose it has at least one whole root. Then he is a divider of the number 2 (Member without X). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Substitute x \u003d -1 :
.

So, we found one root X \u003d -1 . We divide on X - (-1) \u003d x + 1:


So,
.

Now you need to solve the third degree equation:
.
If we assume that this equation has a whole root, then it is a divider of the number 2 (Member without X). That is, the whole root can be one of the numbers:
1, 2, -1, -2 .
Substitute x \u003d -1 :
.

So, we found another root X \u003d -1 . It would be possible, as in the previous case, divide the polynomial to, but we grouped members:
.

Since equation X. 2 + 2 = 0 It does not have real roots, we received a decomposition of the denominator for factors:
.

2. We decompose the fraction on the simplest. We are looking for a decomposition in the form:
.
We are freed from the denominator the fractions, multiply on (x + 1) 2 (x 2 + 2):
(3.1) .
Substitute x \u003d -1 . Then X +. 1 = 0 ,
.

Differentiasis (3.1) :

;

.
Substitute x \u003d -1 And we take into account that X + 1 = 0 :
;
; .

Substitute B. (3.1) x \u003d. 0 :
0 \u003d 2 A + 2 B + D;
.

Ensure B. (3.1) Coefficients at X. 3 :
;
1 \u003d B + C;
.

So, we found a decomposition on the simplest fractions:
.

3. We integrate.


.

Topic: integrating rational fractions.

Attention! When studying one of the main integration techniques: integrating rational fractions - is required for strict evidence to consider polynomials in the complex area. Therefore, it is necessary explore pre- some properties of integrated numbers and operations over them.

Integrating the simplest rational fractions.

If a P.(z.) and Q.(z.) - polynomials in the complex region, then - rational fraction. It is called rightif the degree P.(z.) less degree Q.(z.) , I. wrongif the degree R no less than degree Q..

Any incorrect fraction can be represented as: ,

P (z) \u003d q (z) s (z) + r (z),

a. R.(z.) – polynomial, the degree of less degree Q.(z.).

Thus, the integration of rational fractions is reduced to the integration of polynomials, that is, power functions, and the correct fractions, as it is the right fraction.

Definition 5. Simplest (or elementary) fractions are called fractions of the following types:

1) , 2) , 3) , 4) .

Throw out how they integrate.

3) (studied earlier).

Theorem 5. All correct fraction can be represented as the sum of the simplest fractions (without proof).

Corollary 1. If it is the right rational fraction, and if only simple valid roots are among the roots of the numerous, then only the simplest fractions of the 1st type will be present in the expansion of the fraction on the sum of the simplest fractions:

Example 1.

Corollary 2. If it is the right rational fraction, and if only multiple valid roots are among the roots, only the simplest fractions of the 1st and 2nd types will be present in the decomposition of the shortest fractions:

Example 2.

Corollary 3. If there is a correct rational fraction, and if there will be only simple complex - conjugate roots among the roots of the polynomial, then only the simplest fractions of the 3rd type will be present in the expansion of the protrusion of the simplest fractions:

Example 3.

Corollary 4. If there is a correct rational fraction, and if only multiple complex roots are polynomial among the roots of the polynomial, then only the simplest fractions of the 3rd and 4th types will be present in the expansion of the shortest fractions:

To determine unknown coefficients in the shown decompositions are applied as follows. The left and right-hand part of the decomposition containing unknown coefficients multiply on the equality of two polynomials. From it, the equations are obtained on the desired coefficients using that:

1. Equality is valid for any x (private values). In this case, it turns out how many equations, any M of which allow you to find unknown coefficients.

2. Coefficients are coincided with the same degrees X (method of indefinite coefficients). In this case, the M system is obtained by M - unknown, from which unknown coefficients are found.

3. Combined method.

Example 5. Dispatch fraction on the simplest.

Decision:

Find the coefficients A and V.

1 method - method of private values:

2 Method - Method of uncertain coefficients:

Answer:

Integrating rational fractions.

Theorem 6. An indefinite integral from any rational fraction at any interval on which its denominator is not equal to zero., exists and expressed through elementary functions, namely rational fractions, logarithms and arctshantes.

Evidence.

Imagine a rational fraction in the form: . In this case, the last term is the right shot, and in the theorem 5 it can be represented as a linear combination of the simplest fractions. Thus, the integration of rational fraction is reduced to the integration of the polynomial S.(x.) and the simplest fractions, whose primitive, as shown, have the appearance specified in the theorem.

Comment. The main difficulty is the decomposition of the denominator for multipliers, that is, the search for all its roots.

Example 1. Find an integral

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

1. If the rational fraction is incorrect, it is represented as the sum of the polynomial and the correct rational fraction (see paragraph 2).

In this particular, the integration of the wrong rational fraction is reduced to the integration of the polynomial and the correct rational fraction.

2. decompose the denominator of the correct fraction on multipliers.

3. The correct rational fraction is decomposed on the sum of the simplest fractions. This is the most integration of the correct rational fraction to integrate the simplest fractions.

Consider examples.

Example 1. Find.

Decision. Under the integral is an incorrect rational fraction. Having highlighting the whole part

Hence,

Noticing that, spread the right rational fraction

on the simplest fractions:

(see formula (18)). therefore

Thus, we finally have

Example 2. Find

Decision. Under the integral is the right rational fraction.

Decompose it on the simplest fraction (see formula (16)), we get