The derivation of formulas for calculating integrals of the simplest, elementary, fractions of four types is given. More complex integrals, of the fourth type of fractions, are calculated using the reduction formula. An example of integrating a fraction of the fourth type is considered.

Content

See also: Indefinite Integral Table
Methods for calculating indefinite integrals

As you know, any rational function of some variable x can be decomposed into a polynomial and the simplest, elementary, fractions. There are four types of simple fractions:
1) ;
2) ;
3) ;
4) .
Here a, A, B, b, c are real numbers. Equation x 2 + bx + c \u003d 0 has no valid roots.

Integration of fractions of the first two types

Integration of the first two fractions is performed using the following formulas from the table of integrals:
,
, n ≠ - 1 .

1. Integration of a fraction of the first type

The fraction of the first type is reduced by the substitution t \u003d x - a to a tabular integral:
.

2. Integration of a fraction of the second type

A fraction of the second type is reduced to a tabular integral by the same substitution t \u003d x - a:

.

3. Integration of fraction of the third type

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

3.1. Step 1. Select the derivative of the denominator in the numerator

Select the derivative of the denominator in the numerator of the fraction. We denote: u \u003d x 2 + bx + c... Differentiate: u ′ \u003d 2 x + b... Then
;
.
But
.
We have omitted the modulus 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 bring the denominator of the fraction to the sum of squares:
,
where.
We assume that the equation x 2 + bx + c \u003d 0 has no roots. Therefore .

Let's make a substitution
,
.
.

So,
.

Thus, we have found the integral of the third type of fraction:

,
where.

4. Integration of the fourth type fraction

Finally, consider the integral of the fourth type of fraction:
.
We calculate it in three steps.

4.1) Select the derivative of the denominator in the numerator:
.

4.2) Calculate the integral
.

4.3) Calculate the integrals
,
using the cast formula:
.

4.1. Step 1. Selecting the derivative of the denominator in the numerator

Select the derivative of the denominator in the numerator, as we did in. We denote u \u003d x 2 + bx + c... Differentiate: u ′ \u003d 2 x + b... Then
.

.
But
.

Finally, we have:
.

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

We calculate the integral
.
Its calculation is outlined in.

4.3. Step 3. Derivation of the reduction formula

Now consider the integral
.

We bring the square trinomial to the sum of squares:
.
Here .
We make a substitution.
.
.

We perform transformations and integrate piece by piece.




.

Multiply by 2 (n - 1):
.
Back to x and I n.
,
;
;
.

So, for I n we got the reduction formula:
.
Applying this formula successively, we reduce the integral I n to I 1 .

Example

Calculate the integral

1. Select the derivative of the denominator in the numerator.
;
;


.
Here
.

2. We calculate the integral of the simplest fraction.

.

3. We apply the reduction formula:

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

.

Finally, we have:

.
We find the coefficient at.
.

See also:

The material presented in this topic is based on the information presented in the topic "Rational fractions. Decomposition of rational fractions into elementary (simplest) fractions". I strongly advise you to at least briefly review this topic before proceeding to reading this material. In addition, we will need a table of indefinite integrals.

Let me remind you of a couple of terms. They were discussed in the relevant topic, therefore I will limit myself here to a brief wording.

The ratio of two polynomials $ \\ frac (P_n (x)) (Q_m (x)) $ is called a rational function or a rational fraction. The rational fraction is called correctif $ n< m$, т.е. если степень многочлена, стоящего в числителе, меньше степени многочлена, стоящего в знаменателе. В противном случае (если $n ≥ m$) дробь называется wrong.

Elementary (simplest) rational fractions are called rational fractions of four types:

1. $ \\ frac (A) (x-a) $;
2. $ \\ frac (A) ((x-a) ^ n) $ ($ n \u003d 2,3,4, \\ ldots $);
3. $ \\ frac (Mx + N) (x ^ 2 + px + q) $ ($ p ^ 2-4q< 0$);
4. $ \\ frac (Mx + N) ((x ^ 2 + px + q) ^ n) $ ($ p ^ 2-4q< 0$; $n=2,3,4,\ldots$).

Note (desirable for a more complete understanding of the text): show / hide

Why is the condition $ p ^ 2-4q needed< 0$ в дробях третьего и четвертого типов? Рассмотрим квадратное уравнение $x^2+px+q=0$. Дискриминант этого уравнения $D=p^2-4q$. По сути, условие $p^2-4q < 0$ означает, что $D < 0$. Если $D < 0$, то уравнение $x^2+px+q=0$ не имеет действительных корней. Т.е. выражение $x^2+px+q$ неразложимо на множители. Именно эта неразложимость нас и интересует.

For example, for the expression $ x ^ 2 + 5x + 10 $ we get: $ p ^ 2-4q \u003d 5 ^ 2-4 \\ cdot 10 \u003d -15 $. Since $ p ^ 2-4q \u003d -15< 0$, то выражение $x^2+5x+10$ нельзя разложить на множители.

By the way, for this check it is not at all necessary that the coefficient in front of $ x ^ 2 $ equals 1. For example, for $ 5x ^ 2 + 7x-3 \u003d 0 $ we get: $ D \u003d 7 ^ 2-4 \\ cdot 5 \\ cdot (-3) \u003d 109 $. Since $ D\u003e 0 $, the expression $ 5x ^ 2 + 7x-3 $ can be factorized.

Examples of rational fractions (right and wrong), as well as examples of decomposing rational fractions into elementary ones can be found. Here we will be interested only in the issues of their integration. Let's start by integrating elementary fractions. So, each of the four types of the above elementary fractions is easy to integrate using the formulas below. Let me remind you that when integrating fractions of the type (2) and (4), it is assumed that $ n \u003d 2,3,4, \\ ldots $. Formulas (3) and (4) require the fulfillment of the condition $ p ^ 2-4q< 0$.

\\ begin (equation) \\ int \\ frac (A) (xa) dx \u003d A \\ cdot \\ ln | xa | + C \\ end (equation) \\ begin (equation) \\ int \\ frac (A) ((xa) ^ n ) dx \u003d - \\ frac (A) ((n-1) (xa) ^ (n-1)) + C \\ end (equation) \\ begin (equation) \\ int \\ frac (Mx + N) (x ^ 2 + px + q) dx \u003d \\ frac (M) (2) \\ cdot \\ ln (x ^ 2 + px + q) + \\ frac (2N-Mp) (\\ sqrt (4q-p ^ 2)) \\ arctg \\ For $ \\ int \\ frac (Mx + N) ((x ^ 2 + px + q) ^ n) dx $, replace $ t \u003d x + \\ frac (p) (2) $, after which the resulting integral is split into two. The first will be calculated by entering the differential under the sign, and the second will have the form $ I_n \u003d \\ int \\ frac (dt) ((t ^ 2 + a ^ 2) ^ n) $. This integral is taken using the recurrence relation

\\ begin (equation) I_ (n + 1) \u003d \\ frac (1) (2na ^ 2) \\ frac (t) ((t ^ 2 + a ^ 2) ^ n) + \\ frac (2n-1) (2na ^ 2) I_n, \\; n \\ in N \\ end (equation)

The calculation of such an integral is discussed in example №7 (see the third part).

Scheme for calculating integrals of rational functions (rational fractions):

If the integrand is elementary, then apply formulas (1) - (4).

1. If the integrand is not elementary, then represent it as a sum of elementary fractions, and then integrate using formulas (1) - (4).
2. The above algorithm for integrating rational fractions has an undeniable merit - it is universal. Those. using this algorithm one can integrate

any {!LANG-0fcc8da2129973949fcbd55fcca720ed!} rational fraction. That is why almost all changes of variables in an indefinite integral (substitutions of Euler, Chebyshev, universal trigonometric substitution) are done in such a way that after this change we get a rational fraction under the integral. And to it already apply the algorithm. We will analyze the direct application of this algorithm by examples, after making a small note.

$$ \\ int \\ frac (7dx) (x + 9) \u003d 7 \\ ln | x + 9 | + C. $$

In principle, this integral is not difficult to obtain without mechanical application of the formula. If we take the constant $ 7 $ out of the integral sign and take into account that $ dx \u003d d (x + 9) $, then we get:

$$ \\ int \\ frac (7dx) (x + 9) \u003d 7 \\ cdot \\ int \\ frac (dx) (x + 9) \u003d 7 \\ cdot \\ int \\ frac (d (x + 9)) (x + 9 ) \u003d | u \u003d x + 9 | \u003d 7 \\ cdot \\ int \\ frac (du) (u) \u003d 7 \\ ln | u | + C \u003d 7 \\ ln | x + 9 | + C. $$

For detailed information, I recommend looking at the topic. It explains in detail how such integrals are solved. By the way, the formula is proved by the same transformations that were applied in this paragraph when solving "manually".

2) Again, there are two ways: apply the ready-made formula or do without it. If you apply the formula, then you should take into account that the coefficient in front of $ x $ (number 4) will have to be removed. To do this, we simply put the four out of brackets:

$$ \\ int \\ frac (11dx) ((4x + 19) ^ 8) \u003d \\ int \\ frac (11dx) (\\ left (4 \\ left (x + \\ frac (19) (4) \\ right) \\ right) ^ 8) \u003d \\ int \\ frac (11dx) (4 ^ 8 \\ left (x + \\ frac (19) (4) \\ right) ^ 8) \u003d \\ int \\ frac (\\ frac (11) (4 ^ 8) dx) (\\ left (x + \\ frac (19) (4) \\ right) ^ 8). $$

Now it's time to apply the formula:

$$ \\ int \\ frac (\\ frac (11) (4 ^ 8) dx) (\\ left (x + \\ frac (19) (4) \\ right) ^ 8) \u003d - \\ frac (\\ frac (11) (4 ^ 8)) ((8-1) \\ left (x + \\ frac (19) (4) \\ right) ^ (8-1)) + C \u003d - \\ frac (\\ frac (11) (4 ^ 8)) (7 \\ left (x + \\ frac (19) (4) \\ right) ^ 7) + C \u003d - \\ frac (11) (7 \\ cdot 4 ^ 8 \\ left (x + \\ frac (19) (4) \\ right ) ^ 7) + C. $$

You can do without using the formula. And even without putting the $ 4 $ constant in parentheses. Taking into account that $ dx \u003d \\ frac (1) (4) d (4x + 19) $, we get:

$$ \\ int \\ frac (11dx) ((4x + 19) ^ 8) \u003d 11 \\ int \\ frac (dx) ((4x + 19) ^ 8) \u003d \\ frac (11) (4) \\ int \\ frac ( d (4x + 19)) ((4x + 19) ^ 8) \u003d | u \u003d 4x + 19 | \u003d \\\\ \u003d \\ frac (11) (4) \\ int \\ frac (du) (u ^ 8) \u003d \\ \\\\ \u003d \\ frac (11) (4) \\ cdot \\ frac (u ^ (- 7)) (- 7) + C \u003d - \\ frac (11) (28) \\ cdot \\ frac (1) (u ^ 7 ) + C \u003d - \\ frac (11) (28 (4x + 19) ^ 7) + C. $$

Detailed explanations on finding such integrals are given in the topic "Integration by substitution (entering under the differential sign)".

3) We need to integrate the fraction $ \\ frac (4x + 7) (x ^ 2 + 10x + 34) $. This fraction has the structure $ \\ frac (Mx + N) (x ^ 2 + px + q) $, where $ M \u003d 4 $, $ N \u003d 7 $, $ p \u003d 10 $, $ q \u003d 34 $. However, to make sure that this is really an elementary fraction of the third type, you need to check the condition $ p ^ 2-4q< 0$. Так как $p^2-4q=10^2-4\cdot 34=-16 < 0$, то мы действительно имеем дело с интегрированием элементарной дроби третьего типа. Как и в предыдущих пунктах есть два пути для нахождения $\int\frac{4x+7}{x^2+10x+34}dx$. Первый путь - банально использовать формулу . Подставив в неё $M=4$, $N=7$, $p=10$, $q=34$ получим:

$$ \\ int \\ frac (4x + 7) (x ^ 2 + 10x + 34) dx \u003d \\ frac (4) (2) \\ cdot \\ ln (x ^ 2 + 10x + 34) + \\ frac (2 \\ cdot 7-4 \\ cdot 10) (\\ sqrt (4 \\ cdot 34-10 ^ 2)) \\ arctg \\ frac (2x + 10) (\\ sqrt (4 \\ cdot 34-10 ^ 2)) + C \u003d \\\\ \u003d 2 \\ cdot \\ ln (x ^ 2 + 10x + 34) + \\ frac (-26) (\\ sqrt (36)) \\ arctg \\ frac (2x + 10) (\\ sqrt (36)) + C \u003d 2 \\ cdot \\ ln (x ^ 2 + 10x + 34) + \\ frac (-26) (6) \\ arctg \\ frac (2x + 10) (6) + C \u003d \\\\ \u003d 2 \\ cdot \\ ln (x ^ 2 + 10x +34) - \\ frac (13) (3) \\ arctg \\ frac (x + 5) (3) + C. $$

Let's solve the same example, but without using the ready-made formula. Let's try to select the derivative of the denominator in the numerator. What does this mean? We know that $ (x ^ 2 + 10x + 34) "\u003d 2x + 10 $. It is the expression $ 2x + 10 $ that we have to isolate in the numerator. So far, the numerator contains only $ 4x + 7 $, but this will not be long. Let's apply the following transformation to the numerator:

$$ 4x + 7 \u003d 2 \\ cdot 2x + 7 \u003d 2 \\ cdot (2x + 10-10) + 7 \u003d 2 \\ cdot (2x + 10) -2 \\ cdot 10 + 7 \u003d 2 \\ cdot (2x + 10) -thirteen. $$

Now the required expression $ 2x + 10 $ appears in the numerator. And our integral can be rewritten as follows:

$$ \\ int \\ frac (4x + 7) (x ^ 2 + 10x + 34) dx \u003d \\ int \\ frac (2 \\ cdot (2x + 10) -13) (x ^ 2 + 10x + 34) dx. $$

Let's split the integrand into two. Well, and, accordingly, the integral itself is also "bifurcated":

$$ \\ int \\ frac (2 \\ cdot (2x + 10) -13) (x ^ 2 + 10x + 34) dx \u003d \\ int \\ left (\\ frac (2 \\ cdot (2x + 10)) (x ^ 2 + 10x + 34) - \\ frac (13) (x ^ 2 + 10x + 34) \\ right) \\; dx \u003d \\\\ \u003d \\ int \\ frac (2 \\ cdot (2x + 10)) (x ^ 2 + 10x + 34) dx- \\ int \\ frac (13dx) (x ^ 2 + 10x + 34) \u003d 2 \\ cdot \\ int \\ frac ((2x + 10) dx) (x ^ 2 + 10x + 34) -13 \\ cdot \\ int \\ frac (dx) (x ^ 2 + 10x + 34). $$

Let's talk first about the first integral, i.e. about $ \\ int \\ frac ((2x + 10) dx) (x ^ 2 + 10x + 34) $. Since $ d (x ^ 2 + 10x + 34) \u003d (x ^ 2 + 10x + 34) "dx \u003d (2x + 10) dx $, the denominator differential is located in the numerator of the integrand. In short, instead of the expression $ ( 2x + 10) dx $ we write $ d (x ^ 2 + 10x + 34) $.

Now let's say a few words about the second integral. Allocate a complete square in the denominator: $ x ^ 2 + 10x + 34 \u003d (x + 5) ^ 2 + 9 $. In addition, we take into account $ dx \u003d d (x + 5) $. Now the sum of integrals we obtained earlier can be rewritten in a slightly different form:

$$ 2 \\ cdot \\ int \\ frac ((2x + 10) dx) (x ^ 2 + 10x + 34) -13 \\ cdot \\ int \\ frac (dx) (x ^ 2 + 10x + 34) \u003d 2 \\ cdot \\ int \\ frac (d (x ^ 2 + 10x + 34)) (x ^ 2 + 10x + 34) -13 \\ cdot \\ int \\ frac (d (x + 5)) ((x + 5) ^ 2 + 9). $$

If in the first integral we make the replacement $ u \u003d x ^ 2 + 10x + 34 $, then it will take the form $ \\ int \\ frac (du) (u) $ and is taken by a simple application of the second formula from. As for the second integral, the replacement $ u \u003d x + 5 $ is realizable for it, after which it takes the form $ \\ int \\ frac (du) (u ^ 2 + 9) $. This is the purest eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we will have:

$$ 2 \\ cdot \\ int \\ frac (d (x ^ 2 + 10x + 34)) (x ^ 2 + 10x + 34) -13 \\ cdot \\ int \\ frac (d (x + 5)) ((x + 5) ^ 2 + 9) \u003d 2 \\ cdot \\ ln (x ^ 2 + 10x + 34) - \\ frac (13) (3) \\ arctg \\ frac (x + 5) (3) + C. $$

We got the same answer as when applying the formula, which, strictly speaking, is not surprising. In general, the formula is proved by the same methods that we used to find this integral. I believe that an attentive reader may have one question here, so I will formulate it:

Question number 1

If we apply the second formula from the table of indefinite integrals to the integral $ \\ int \\ frac (d (x ^ 2 + 10x + 34)) (x ^ 2 + 10x + 34) $, then we get the following:

$$ \\ int \\ frac (d (x ^ 2 + 10x + 34)) (x ^ 2 + 10x + 34) \u003d | u \u003d x ^ 2 + 10x + 34 | \u003d \\ int \\ frac (du) (u) \u003d \\ ln | u | + C \u003d \\ ln | x ^ 2 + 10x + 34 | + C. $$

Why was the module missing in the solution?

Answer to question number 1

The question is completely natural. The module was absent only because the expression $ x ^ 2 + 10x + 34 $ for any $ x \\ in R $ is greater than zero. This is quite easy to show in several ways. For example, since $ x ^ 2 + 10x + 34 \u003d (x + 5) ^ 2 + 9 $ and $ (x + 5) ^ 2 ≥ 0 $, then $ (x + 5) ^ 2 + 9\u003e 0 $ ... You can think differently, without involving the selection of a complete square. Since $ 10 ^ 2-4 \\ cdot 34 \u003d -16< 0$, то $x^2+10x+34 > 0 $ for any $ x \\ in R $ (if this logical chain is surprising, I advise you to look at the graphical method for solving square inequalities). In any case, since $ x ^ 2 + 10x + 34\u003e 0 $, then $ | x ^ 2 + 10x + 34 | \u003d x ^ 2 + 10x + 34 $, i.e. you can use regular brackets instead of a module.

All points of example No. 1 are solved, it remains only to write down the answer.

Answer:

1. $ \\ int \\ frac (7dx) (x + 9) \u003d 7 \\ ln | x + 9 | + C $;
2. $ \\ int \\ frac (11dx) ((4x + 19) ^ 8) \u003d - \\ frac (11) (28 (4x + 19) ^ 7) + C $;
3. $ \\ int \\ frac (4x + 7) (x ^ 2 + 10x + 34) dx \u003d 2 \\ cdot \\ ln (x ^ 2 + 10x + 34) - \\ frac (13) (3) \\ arctg \\ frac (x +5) (3) + C $.

Example No. 2

Find the integral $ \\ int \\ frac (7x + 12) (3x ^ 2-5x-2) dx $.

At first glance, the integrand $ \\ frac (7x + 12) (3x ^ 2-5x-2) $ is very similar to the elementary fraction of the third type, i.e. by $ \\ frac (Mx + N) (x ^ 2 + px + q) $. It seems that the only difference is the $ 3 $ coefficient in front of $ x ^ 2 $, but after all, it won't take long to remove the coefficient (put out the brackets). However, this similarity is apparent. For the fraction $ \\ frac (Mx + N) (x ^ 2 + px + q) $, the condition $ p ^ 2-4q< 0$, которое гарантирует, что знаменатель $x^2+px+q$ нельзя разложить на множители. Проверим, как обстоит дело с разложением на множители у знаменателя нашей дроби, т.е. у многочлена $3x^2-5x-2$.

Our coefficient in front of $ x ^ 2 $ is not equal to one, therefore, check the condition $ p ^ 2-4q< 0$ напрямую мы не можем. Однако тут нужно вспомнить, откуда взялось выражение $p^2-4q$. Это всего лишь дискриминант квадратного уравнения $x^2+px+q=0$. Если дискриминант меньше нуля, то выражение $x^2+px+q$ на множители не разложишь. Вычислим дискриминант многочлена $3x^2-5x-2$, расположенного в знаменателе нашей дроби: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. Итак, $D > 0 $, so the expression $ 3x ^ 2-5x-2 $ can be factorized. This means that the fraction $ \\ frac (7x + 12) (3x ^ 2-5x-2) $ is not an elementary fraction of the third type, and apply to the integral $ \\ int \\ frac (7x + 12) (3x ^ 2- 5x-2) dx $ formula is not allowed.

Well, if a given rational fraction is not elementary, then it must be represented as a sum of elementary fractions, and then integrated. In short, take advantage of the trail. How to decompose a rational fraction into elementary ones is written in detail. Let's start by factoring the denominator:

$$ 3x ^ 2-5x-2 \u003d 0; \\\\ \\ begin (aligned) & D \u003d (- 5) ^ 2-4 \\ cdot 3 \\ cdot (-2) \u003d 49; \\\\ & x_1 \u003d \\ frac ( - (- 5) - \\ sqrt (49)) (2 \\ cdot 3) \u003d \\ frac (5-7) (6) \u003d \\ frac (-2) (6) \u003d - \\ frac (1) (3); \\\\ & x_2 \u003d \\ frac (- (- 5) + \\ sqrt (49)) (2 \\ cdot 3) \u003d \\ frac (5 + 7) (6) \u003d \\ frac (12) (6) \u003d 2. \\ 3 \\ cdot \\ left (x + \\ frac (1) (3) \\ right) (x-2). $$

We represent the subinteral fraction as follows:

$$ \\ frac (7x + 12) (3x ^ 2-5x-2) \u003d \\ frac (7x + 12) (3 \\ cdot \\ left (x + \\ frac (1) (3) \\ right) (x-2) ) \u003d \\ frac (\\ frac (7) (3) x + 4) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)). $$

Now we expand the fraction $ \\ frac (\\ frac (7) (3) x + 4) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)) $ into elementary ones:

$$ \\ frac (\\ frac (7) (3) x + 4) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)) \u003d \\ frac (A) (x + \\ frac ( 1) (3)) + \\ frac (B) (x-2) \u003d \\ frac (A (x-2) + B \\ left (x + \\ frac (1) (3) \\ right)) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)); \\\\ \\ frac (7) (3) x + 4 \u003d A (x-2) + B \\ left (x + \\ frac (1) ( 3) \\ right). $$

There are two standard ways to find the coefficients of $ A $ and $ B $: the undefined coefficient method and the partial value substitution method. Let's apply the partial value substitution method by substituting $ x \u003d 2 $ and then $ x \u003d - \\ frac (1) (3) $:

$$ \\ frac (7) (3) x + 4 \u003d A (x-2) + B \\ left (x + \\ frac (1) (3) \\ right). \\\\ x \u003d 2; \\; \\ frac (7) (3) \\ cdot 2 + 4 \u003d A (2-2) + B \\ left (2+ \\ frac (1) (3) \\ right); \\; \\ frac (26) (3) \u003d \\ frac (7) (3) B; \\; B \u003d \\ frac (26) (7). \\\\ x \u003d - \\ frac (1) (3); \\; \\ frac (7) (3) \\ cdot \\ left (- \\ frac (1) (3) \\ right) + 4 \u003d A \\ left (- \\ frac (1) (3) -2 \\ right) + B \\ left (- \\ frac (1) (3) + \\ frac (1) (3) \\ right); \\; \\ frac (29) (9) \u003d - \\ frac (7) (3) A; \\; A \u003d - \\ frac (29 \\ cdot 3) (9 \\ cdot 7) \u003d - \\ frac (29) (21). \\\\ $$

Since the coefficients have been found, it remains only to write down the finished decomposition:

$$ \\ frac (\\ frac (7) (3) x + 4) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)) \u003d \\ frac (- \\ frac (29) ( 21)) (x + \\ frac (1) (3)) + \\ frac (\\ frac (26) (7)) (x-2). $$

In principle, you can leave such a record, but I prefer a more accurate option:

$$ \\ frac (\\ frac (7) (3) x + 4) (\\ left (x + \\ frac (1) (3) \\ right) (x-2)) \u003d - \\ frac (29) (21) \\ $$

Returning to the original integral, we substitute the resulting expansion into it. Then we split the integral into two, and apply the formula to each. I prefer to move the constants outside the integral sign right away:

$$ \\ int \\ frac (7x + 12) (3x ^ 2-5x-2) dx \u003d \\ int \\ left (- \\ frac (29) (21) \\ cdot \\ frac (1) (x + \\ frac (1) (3)) + \\ frac (26) (7) \\ cdot \\ frac (1) (x-2) \\ right) dx \u003d \\\\ \u003d \\ int \\ left (- \\ frac (29) (21) \\ cdot \\ \\ frac (29) (21) \\ cdot \\ int \\ frac (dx) (x + \\ frac (1) (3)) + \\ frac (26) (7) \\ cdot \\ int \\ frac (dx) (x-2 ) dx \u003d \\\\ \u003d - \\ frac (29) (21) \\ cdot \\ ln \\ left | x + \\ frac (1) (3) \\ right | + \\ frac (26) (7) \\ cdot \\ ln | x- 2 | + C. $$

Answer: $ \\ int \\ frac (7x + 12) (3x ^ 2-5x-2) dx \u003d - \\ frac (29) (21) \\ cdot \\ ln \\ left | x + \\ frac (1) (3) \\ right | + \\ frac (26) (7) \\ cdot \\ ln | x-2 | + C $.

Example No. 3

Find the integral $ \\ int \\ frac (x ^ 2-38x + 157) ((x-1) (x + 4) (x-9)) dx $.

We need to integrate the fraction $ \\ frac (x ^ 2-38x + 157) ((x-1) (x + 4) (x-9)) $. The numerator contains a polynomial of the second degree, and the denominator contains a polynomial of the third degree. Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. $ 2< 3$, то подынтегральная дробь является правильной. Разложение этой дроби на элементарные (простейшие) было получено в примере №3 на странице, посвящённой разложению рациональных дробей на элементарные. Полученное разложение таково:

$$ \\ frac (x ^ 2-38x + 157) ((x-1) (x + 4) (x-9)) \u003d - \\ frac (3) (x-1) + \\ frac (5) (x +4) - \\ frac (1) (x-9). $$

We just have to split the given integral into three, and apply the formula to each. I prefer to immediately move constants outside the integral sign:

$$ \\ int \\ frac (x ^ 2-38x + 157) ((x-1) (x + 4) (x-9)) dx \u003d \\ int \\ left (- \\ frac (3) (x-1) + \\ frac (5) (x + 4) - \\ frac (1) (x-9) \\ right) dx \u003d \\\\ \u003d - 3 \\ cdot \\ int \\ frac (dx) (x-1) + 5 \\ cdot \\ int \\ frac (dx) (x + 4) - \\ int \\ frac (dx) (x-9) \u003d - 3 \\ ln | x-1 | +5 \\ ln | x + 4 | - \\ ln | x- 9 | + C. $$

Answer: $ \\ int \\ frac (x ^ 2-38x + 157) ((x-1) (x + 4) (x-9)) dx \u003d -3 \\ ln | x-1 | +5 \\ ln | x + 4 | - \\ ln | x-9 | + C $.

The continuation of the analysis of examples of this topic is located in the second part.

The fraction is called correctif the highest degree of the numerator is less than the highest degree of the denominator. The integral of a regular rational fraction is:

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

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

1. Only complex roots, then it is necessary to select a complete square from it: $$ \\ int \\ frac (mx + n) (ax ^ 2 + bx + c) dx \u003d \\ int \\ frac (mx + n) (x ^ 2 \\ pm a ^ 2) $$
2. Different real roots $ x_1 $ and $ x_2 $, then you need to expand the integral and find the undefined coefficients $ A $ and $ B $: $$ \\ int \\ frac (mx + n) (ax ^ 2 + bx + c) dx \u003d \\ int \\ frac (A) (x-x_1) dx + \\ int \\ frac (B) (x-x_2) dx $$
3. One multiple root $ x_1 $, then we expand the integral and find the undefined coefficients $ A $ and $ B $ for the following formula: $$ \\ int \\ frac (mx + n) (ax ^ 2 + bx + c) dx \u003d \\ int \\ frac (A) ((x-x_1) ^ 2) dx + \\ int \\ frac (B) (x-x_1) dx $$

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 the rational fraction is:

$$ \\ int \\ frac (P (x)) (ax ^ 2 + bx + c) dx \u003d \\ int Q (x) dx + \\ int \\ frac (mx + n) (ax ^ 2 + bx + c) dx $$

Solution examples

Example 1

Find the Integral of the Rational Fraction: $$ \\ int \\ frac (dx) (x ^ 2-10x + 16) $$

Decision

The fraction is regular and the polynomial has only complex roots. Therefore, let's select a complete square: $$ \\ int \\ frac (dx) (x ^ 2-10x + 16) \u003d \\ int \\ frac (dx) (x ^ 2-2 \\ cdot 5 x + 5 ^ 2 - 9) \u003d $$ We fold a complete square and bring it under the differential sign $ x-5 $: $$ \u003d \\ int \\ frac (dx) ((x-5) ^ 2 - 9) \u003d \\ int \\ frac (d (x-5)) ((x-5) ^ 2-9) \u003d $$ Using the table of integrals we get: $$ \u003d \\ frac (1) (2 \\ cdot 3) \\ ln \\ bigg | \\ frac (x-5 - 3) (x-5 + 3) \\ bigg | + C \u003d \\ frac (1) (6) \\ ln \\ bigg | \\ frac (x-8) (x-2) \\ bigg | + C $$ If you can't solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the course of the calculation and get information. This will help you get credit from the teacher in a timely manner!

Answer

$$ \\ int \\ frac (dx) (x ^ 2-10x + 16) \u003d \\ frac (1) (6) \\ ln \\ bigg | \\ frac (x-8) (x-2) \\ bigg | + C $$

Example 2

Integrate rational fractions: $$ \\ int \\ frac (x + 2) (x ^ 2 + 5x-6) dx $$

Decision

Solve the quadratic equation: $$ x ^ 2 + 5x-6 \u003d 0 $$ $$ x_ (12) \u003d \\ frac (-5 \\ pm \\ sqrt (25-4 \\ cdot 1 \\ cdot (-6))) (2) \u003d \\ frac (-5 \\ pm 7) (2) $$ We write down the roots: $$ x_1 \u003d \\ frac (-5-7) (2) \u003d -6; x_2 \u003d \\ frac (-5 + 7) (2) \u003d 1 $$ Taking into account the obtained roots, we transform the integral: $$ \\ int \\ frac (x + 2) (x ^ 2 + 5x-6) dx \u003d \\ int \\ frac (x + 2) ((x-1) (x + 6)) dx \u003d $$ We perform the expansion of the rational fraction: $$ \\ frac (x + 2) ((x-1) (x + 6)) \u003d \\ frac (A) (x-1) + \\ frac (B) (x + 6) \u003d \\ frac (A (x -6) + B (x-1)) ((x-1) (x + 6)) $$ We equate the numerators and find the coefficients $ A $ and $ B $: $$ A (x + 6) + B (x-1) \u003d x + 2 $$ $$ Ax + 6A + Bx - B \u003d x + 2 $$ $$ \\ begin (cases) A \u200b\u200b+ B \u003d 1 \\\\ 6A - B \u003d 2 \\ end (cases) $$ $$ \\ begin (cases) A \u200b\u200b\u003d \\ frac (3) (7) \\\\ B \u003d \\ frac (4) (7) \\ end (cases) $$ We substitute the found coefficients into the integral and solve it: $$ \\ int \\ frac (x + 2) ((x-1) (x + 6)) dx \u003d \\ int \\ frac (\\ frac (3) (7)) (x-1) dx + \\ int \\ frac (\\ frac (4) (7)) (x + 6) dx \u003d $$ $$ \u003d \\ frac (3) (7) \\ int \\ frac (dx) (x-1) + \\ frac (4) (7) \\ int \\ frac (dx) (x + 6) \u003d \\ frac (3) (7) \\ ln | x-1 | + \\ frac (4) (7) \\ ln | x + 6 | + C $$

Answer

$$ \\ int \\ frac (x + 2) (x ^ 2 + 5x-6) dx \u003d \\ frac (3) (7) \\ ln | x-1 | + \\ frac (4) (7) \\ ln | x + 6 | + C $$

As I have already noted, in integral calculus there is no convenient formula for integrating a fraction. And therefore, there is a sad tendency: the more “sophisticated” the fraction, the more difficult it is to find an integral from it. In this regard, you have to resort to various tricks, which I will now tell you about. Trained readers can immediately benefit from table of contents:

• The method of bringing under the differential sign for the simplest fractions

Artificial Numerator Conversion Method

Example 1

By the way, the considered integral can also be solved by the change of variable method, denoting, but the solution will be written much longer.

Example 2

Find the indefinite integral. Check it out.

This is an example for a do-it-yourself solution. It should be noted that the method of replacing the variable will no longer work here.

Attention, important! Examples Nos. 1,2 are typical and are common... In particular, such integrals often arise in the course of solving other integrals, in particular, when integrating irrational functions (roots).

The considered technique also works in the case if the highest degree of the numerator is greater than the highest degree of the denominator.

Example 3

Find the indefinite integral. Check it out.

We begin to select the numerator.

The algorithm for selecting the numerator is something like this:

1) In the numerator I need to organize, but there. What to do? I put it in brackets and multiply by:.

2) Now I'm trying to open these brackets, what happens? ... Hmm ... it's better, but there is no two in the numerator initially. What to do? You need to multiply by:

3) Expand brackets again:. And here is the first success! The right one turned out! But the problem is that an extra term has appeared. What to do? So that the expression does not change, I must add the same to my construction:
... Life has become easier. Isn't it possible to organize in the numerator again?

4) You can. Trying: ... Expand the brackets of the second term:
... Sorry, but I actually had the previous step, not. What to do? You need to multiply the second term by:

5) Again, for verification, I expand the parentheses in the second term:
... Now it's okay: obtained from the final construction of point 3! But again there is a small "but", an extra term has appeared, which means that I must add to my expression:

If everything is done correctly, then when you expand all the parentheses, we should get the original numerator of the integrand. We check:
Good.

Thus:

Done. In the last term, I applied the method of bringing the function under the differential.

If we find the derivative of the answer and bring the expression to a common denominator, then we get exactly the original integrand. The considered method of decomposition into a sum is nothing but the reverse action to bringing the expression to a common denominator.

The algorithm for selecting the numerator in such examples is best done on a draft. With some skills, it will work out mentally. I recall the record time when I performed a fit for the 11th degree, and the expansion of the numerator took almost two Verd lines.

Example 4

Find the indefinite integral. Check it out.

This is an example for a do-it-yourself solution.

The method of bringing under the differential sign for the simplest fractions

Let's move on to the next type of fractions.
,,, (coefficients and are not equal to zero).

In fact, a couple of cases with arcsine and arctangent have already slipped in the lesson Variable change method in indefinite integral... Such examples are solved by the method of bringing the function under the sign of the differential and further integration using the table. Here are some more typical examples with long and high logarithms:

Example 5

Example 6

Here it is advisable to take the table of integrals in hand and trace by what formulas and as transformation is carried out. Note, how and why the squares are highlighted in these examples. In particular, in example 6, you first need to represent the denominator in the form , then bring it under the differential sign. And all this needs to be done in order to use the standard table formula .

But what to watch, try to solve examples ## 7,8 on your own, especially since they are quite short:

Example 7

Example 8

Find the indefinite integral:

If you can also check these examples, then great respect - your differentiation skills are at their best.

Full square selection method

Integrals of the form, (coefficients and are not equal to zero) are solved full square selection method, which has already appeared in the lesson Geometric transformations of graphs.

In fact, such integrals reduce to one of the four tabular integrals that we have just considered. And this is achieved using the familiar formulas for abbreviated multiplication:

Formulas are applied in this direction, that is, the idea of \u200b\u200bthe method is to artificially organize expressions in the denominator, and then convert them accordingly to either.

Example 9

Find the indefinite integral

This is the simplest example where with the term - unit coefficient (not some number or minus).

We look at the denominator, here the whole thing will obviously come down to a case. Let's start converting the denominator:

Obviously, you need to add 4. And so that the expression does not change, subtract the same four:

Now you can apply the formula:

After the conversion is complete IS ALWAYS it is advisable to perform the reverse move: everything is fine, there are no errors.

The final design of the example in question should look something like this:

Done. Summing up a "free" complex function under the differential sign:, in principle, it could be neglected

Example 10

Find the indefinite integral:

This is an example for a do-it-yourself solution, the answer is at the end of the tutorial.

Example 11

Find the indefinite integral:

What to do when there is a minus in front of it? In this case, you need to take the minus out of the brackets and arrange the terms in the order we need:. Constant ("Two" in this case) do not touch!

Now add one in parentheses. Analyzing the expression, we come to the conclusion that one needs to be one behind the parenthesis - add:

Here we got the formula, we apply:

IS ALWAYS we check on the draft:
, which was required to be verified.

The finishing example looks like this:

Complicating the task

Example 12

Find the indefinite integral:

Here, with the term, it is no longer a unit coefficient, but a "five".

(1) If a constant is found for, then we immediately take it out of the parentheses.

(2) In general, it is always better to take this constant outside the integral so that it does not get in the way under your feet.

(3) Obviously, everything will be reduced to a formula. It is necessary to understand the term, namely, to get a "two"

(4) Yep,. So, we add to the expression, and subtract the same fraction.

(5) Now select a complete square. In the general case, you also need to calculate, but here we have a formula for the long logarithm , and it makes no sense to perform the action, why - it will become clear a little below.

(6) Actually, you can apply the formula , only instead of "x" we have, which does not negate the validity of the tabular integral. Strictly speaking, one step has been omitted - before integration, the function should have been placed under the sign of the differential: but, as I have noted many times, this is often neglected.

(7) In the answer under the root, it is desirable to expand all the parentheses back:

Hard? This is not yet the hardest part in integral calculus. Although, the examples under consideration are not so much complicated as they require good computational techniques.

Example 13

Find the indefinite integral:

This is an example for a do-it-yourself solution. The answer is at the end of the lesson.

There are integrals with roots in the denominator, which, using replacement, reduce to integrals of the considered type, you can read about them in the article Complex integralsbut it is designed for highly prepared students.

Adding the numerator under the differential sign

This is the final part of the lesson, however, integrals of this type are quite common! If fatigue has accumulated, maybe it is better to read it tomorrow? ;)

The integrals that we will consider are similar to the integrals of the previous section, they have the form: or (coefficients, and are not equal to zero).

That is, we have a linear function in the numerator. How to solve such integrals?

Integration of a fractional rational function.
Undefined coefficient method

We continue to integrate fractions. We have already considered 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, you need basic integration skills, 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 deal not so much with finding integrals as ... solving systems of linear equations. In this regard strongly I recommend visiting the lesson Namely - you need to be well versed in the methods of substitution (the "school" method and the method of term-by-term addition (subtraction) of the equations of the system).

What is a fractional rational function? In simple words, a fractional rational function is a fraction, in the numerator and denominator of which there are polynomials or products of polynomials. At the same time, the fractions are more sophisticated than those discussed in the article Integration of some fractions.

Integration of the correct fractional rational function

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

Example 1

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

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

The most significant degree of the numerator is two.

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

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

Output: The highest degree of the numerator STRICTLY less than the leading power of the denominator, which means that the fraction is correct.

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

Now we will consider only regular 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. Factor out the denominator. We look at our denominator:

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

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

General rule: EVERYTHING that can be factorized in the denominator is factorized

We begin to draw up a solution:

Step 3. Using the method of undefined coefficients, we expand the integrand into the sum of simple (elementary) fractions. It will be clearer now.

We look at our integrand function:

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

The question arises, is it possible to do this at all? 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 is only one catch, the odds are we until we do not know, hence the name - the method of undefined coefficients.

As you guessed, the subsequent body movements do not cackle! will be aimed at just to RECOGNIZE them - to find out what they are equal to.

Be careful, I will explain in detail once!

So, we begin to dance from:

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

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

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

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 stone face: In order to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of another polynomial.

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

We compose a system of linear equations.
First, we are looking for the senior degrees:

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

Remember the following nuance well... What would have happened if the right side had not been at all? Say, would it just flaunt 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 assign this very 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 into the second equation of the system:

And finally, mineral water, we select free members.

Eh, ... something I was kidding. All jokes - mathematics is a serious science. In our institute group, no one laughed when the assistant professor said that she would scatter the members along the number line and choose the largest of them. We are in a serious mood. Although ... whoever lives to 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 exactly from the 1st equation, since there smallest odds.

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

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

(4) Substituting into the second (or third) equation, whence we find that

(5) Substituting into the first equation to obtain.

If you have difficulties with the methods of solving the system, practice them in the lesson 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 every equation of the system, as a result everything should "converge".

Almost arrived. The coefficients are found, while:

The finishing job should look something like this:

As you can see, the main difficulty of the assignment was to compose (correctly!) And solve (correctly!) A system of linear equations. And at the final stage, everything is not so complicated: we use the linearity properties of the 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 told about the features of its integration in the lesson Variable change method in indefinite integral.

Check: Differentiate the answer:

The original integrand is obtained, which means that the integral is found correctly.
During the check, it was necessary to bring the expression to a common denominator, and this is not accidental. The method of undefined coefficients and the reduction of an expression to a common denominator are mutually opposite actions.

Example 2

Find the indefinite integral.

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

Example 3

Introduce function

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

Step 2. Can you factor something in the denominator? Obviously not, everything is already laid out. A square trinomial cannot be decomposed into a product for the above reasons. Good. Less work.

Step 3. We represent the fractional rational function as a sum of elementary fractions.
In this case, the decomposition is as follows:

We look at our denominator:
When expanding a fractional-rational function into a sum of elementary fractions, three fundamental points can be distinguished:

1) If the denominator contains a "lone" factor in the first degree (in our case), then we put an indefinite coefficient at the top (in our case). Examples # 1,2 consisted of only such "lonely" multipliers.

2) If the denominator contains multiple multiplier, then you need to decompose 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 multiples: and, take another look at the decomposition I have given and make sure that they are decomposed according to this rule.

3) If the denominator contains an indecomposable polynomial of the second degree (in our case), then in the expansion in the numerator it is necessary to write a linear function with undefined coefficients (in our case, with undefined coefficients and).

In fact, there is also a 4th case, but I will not mention it, since in practice it is extremely rare.

Example 4

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

This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial.
Strictly follow the algorithm!

If you figured out the principles by which you need to expand the 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 you 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 undefined coefficients, we expand the integrand into the sum of elementary fractions:

Please 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 one letter.

We bring the fraction to a common denominator:

Let's compose and solve the system:

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

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

(3) 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. Integration of some fractions.

(3) We use the linearity properties again. In the third integral, we begin to select a complete square (the penultimate paragraph of the lesson Integration of some fractions).

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

(5) Take the third integral. Done.