Primitive fraction. Integration of a fractional-rational function

Enter the function for which you want to find the integral

After calculating the indefinite integral, you can get a DETAILED solution of the integral you entered for free.

Let's find the solution of the indefinite integral of the function f(x) (the antiderivative of the function).

Examples

With the use of degree
(square and cube) and fractions

(x^2 - 1)/(x^3 + 1)

Square root

Sqrt(x)/(x + 1)

cube root

Cbrt(x)/(3*x + 2)

Using sine and cosine

2*sin(x)*cos(x)

Arcsine

X*arcsin(x)

Arc cosine

x*arccos(x)

Application of the logarithm

X*log(x, 10)

natural logarithm

Exhibitor

Tg(x)*sin(x)

Cotangent

Ctg(x)*cos(x)

Irrational fractions

(sqrt(x) - 1)/sqrt(x^2 - x - 1)

Arctangent

X*arctg(x)

Arc tangent

X*arсctg(x)

Hyberbolic sine and cosine

2*sh(x)*ch(x)

Hyberbolic tangent and cotangent

ctgh(x)/tgh(x)

Hyberbolic arcsine and arccosine

X^2*arcsinh(x)*arccosh(x)

Hyberbolic arctangent and arccotangent

X^2*arctgh(x)*arctgh(x)

Rules for entering expressions and functions

Expressions can consist of functions (notations are given in alphabetical order): absolute(x) Absolute value x
(module x or |x|) arccos(x) Function - arc cosine of x arccosh(x) Arc cosine hyperbolic from x arcsin(x) Arcsine from x arcsinh(x) Arcsine hyperbolic from x arctg(x) Function - arc tangent from x arctgh(x) The arc tangent is hyperbolic from x e e a number that is approximately equal to 2.7 exp(x) Function - exponent from x(which is e^x) log(x) or log(x) Natural logarithm of x
(To obtain log7(x), you need to enter log(x)/log(7) (or, for example, for log10(x)=log(x)/log(10)) pi The number is "Pi", which is approximately equal to 3.14 sin(x) Function - Sine of x cos(x) Function - Cosine of x sinh(x) Function - Hyperbolic sine of x cash(x) Function - Hyperbolic cosine of x sqrt(x) The function is the square root of x sqr(x) or x^2 Function - Square x tg(x) Function - Tangent from x tgh(x) Function - Hyperbolic tangent of x cbrt(x) The function is the cube root of x

You can use the following operations in expressions: Real numbers enter in the form 7.5 , not 7,5 2*x- multiplication 3/x- division x^3- exponentiation x + 7- addition x - 6- subtraction
Other features: floor(x) Function - rounding x down (example floor(4.5)==4.0) ceiling(x) Function - rounding x up (example ceiling(4.5)==5.0) sign(x) Function - Sign x erf(x) Error function (or probability integral) laplace(x) Laplace function

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:

1. Only complex roots, then it is necessary to select a full square from it: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(mx+n)(x^2 \pm a ^2) $$
2. Different real roots $ x_1 $ and $ x_2 $, then you need to expand the integral and find the indefinite coefficients $ A $ and $ B $: $$ \int \frac(mx+n)(ax^2+bx+c) dx = \int \frac(A)(x-x_1) dx + \int \frac(B)(x-x_2) dx $$
3. 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

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

See also: Table of indefinite integrals
Methods for calculating indefinite integrals

As is known, any rational function of some variable x can be decomposed into a polynomial and simple, 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=0 has no real roots.

Integration of fractions of the first two types

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

1. Integration of a fraction of the first type

A fraction of the first type by substitution t = x - a is reduced to a table integral:
.

2. Integration of a fraction of the second type

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

.

3. Integration of a fraction of the third type

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

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

We select the derivative of the denominator in the numerator of the fraction. Denote: u = x 2+bx+c. Differentiate: u′ = 2 x + b. Then
;
.
But
.
We omitted the modulo sign because .

Then:
,
where
.

3.2. Step 2. Calculate the integral with A = 0, B=1

Now we calculate the remaining integral:
.

We bring the denominator of the fraction to the sum of squares:
,
where .
We believe that the equation x 2+bx+c=0 has no roots. That's why .

Let's make a substitution
,
.
.

So,
.

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

,
where .

4. Integration of a fraction of the fourth type

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

4.1) We select the derivative of the denominator in the numerator:
.

4.2) Calculate the integral
.

4.3) Calculate integrals
,
using the cast formula:
.

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

We select the derivative of the denominator in the numerator, as we did in . Denote u = x 2+bx+c. Differentiate: u′ = 2 x + b. Then
.

.
But
.

Finally we have:
.

4.2. Step 2. Calculation of the integral with n = 1

We calculate the integral
.
Its calculation is set out 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 by parts.




.

Multiply by 2(n - 1):
.
We return 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 Integral

1. We 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 = 1 , c = 1 , 4 c - b 2 = 3. We write out this formula for n = 2 and n = 3 :
;
.
From here

.

Finally we have:

.
We find the coefficient at .
.

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

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

Thus, the integration of an improper rational fraction is reduced to the integration of a polynomial and a proper rational fraction.

2. Decompose the denominator of a proper fraction into factors.

3. The correct rational fraction is decomposed into the sum of the simplest fractions. Thus, the integration of a proper rational fraction is reduced to the integration of simple fractions.

Consider examples.

Example 1. Find .

Solution. Under the integral is an improper rational fraction. Taking the integer part, we get

Consequently,

Noting that , we expand the proper rational fraction

into simple fractions:

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

Thus, we finally have

Example 2. Find

Solution. Under the integral is a proper rational fraction.

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

The material presented in this topic is based on the information presented in the topic "Rational fractions. Decomposition of rational fractions into elementary (simple) fractions". I strongly advise you to at least skim through 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, so here I will limit myself to a brief formulation.

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 correct if $n< m$, 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 rational fractions of four types:

1. $\frac(A)(x-a)$;
2. $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
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 better understanding of the text): show\hide

Why is the $p^2-4q condition necessary?< 0$ 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: $p^2-4q=5^2-4\cdot 10=-15$. Since $p^2-4q=-15< 0$, то выражение $x^2+5x+10$ нельзя разложить на множители.

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

Examples of rational fractions (regular and improper), as well as examples of the expansion of a rational fraction into elementary ones, can be found. Here we are only interested in questions of their integration. Let's start with the integration of 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 type (2) and (4) $n=2,3,4,\ldots$ is assumed. Formulas (3) and (4) require the condition $p^2-4q< 0$.

\begin(equation) \int \frac(A)(x-a) dx=A\cdot \ln |x-a|+C \end(equation) \begin(equation) \int\frac(A)((x-a)^n )dx=-\frac(A)((n-1)(x-a)^(n-1))+C \end(equation) \begin(equation) \int \frac(Mx+N)(x^2 +px+q) dx= \frac(M)(2)\cdot \ln (x^2+px+q)+\frac(2N-Mp)(\sqrt(4q-p^2))\arctg\ frac(2x+p)(\sqrt(4q-p^2))+C \end(equation)

For $\int\frac(Mx+N)((x^2+px+q)^n)dx$ the replacement $t=x+\frac(p)(2)$ is made, after which the resulting integral is split into two. The first one will be calculated by inserting it under the differential sign, and the second one will look like $I_n=\int\frac(dt)((t^2+a^2)^n)$. This integral is taken using the recurrence relation

\begin(equation) I_(n+1)=\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 analyzed in example No. 7 (see the third part).

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

1. If the integrand is elementary, then apply formulas (1)-(4).
2. If the integrand is not elementary, then represent it as a sum of elementary fractions, and then integrate using formulas (1)-(4).

The above algorithm for integrating rational fractions has an undeniable advantage - it is universal. Those. Using this algorithm, one can integrate any rational fraction. That is why almost all replacements of variables in the indefinite integral (Euler, Chebyshev substitutions, universal trigonometric substitution) are done in such a way that after this replacement we get a rational fraction under the interval. And apply the algorithm to it. We will analyze the direct application of this algorithm using examples, after making a small note.

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

In principle, this integral is easy to obtain without mechanical application of the formula. If we take the constant $7$ out of the integral sign and take into account that $dx=d(x+9)$, then we get:

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

For detailed information I recommend to look 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: to apply a ready-made formula or to do without it. If you apply the formula, then you should take into account that the coefficient in front of $x$ (the number 4) will have to be removed. To do this, we simply take out the four of them in brackets:

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

Now it's time to apply the formula:

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

You can do without using the formula. And even without putting the constant $4$ out of the brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, then we get:

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

Detailed explanations on finding such integrals are given in the topic "Integration by substitution (introduction 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 $M=4$, $N=7$, $p=10$, $q=34$. However, to make sure that this is indeed 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:

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

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

$$ 4x+7=2\cdot 2x+7=2\cdot (2x+10-10)+7=2\cdot(2x+10)-2\cdot 10+7=2\cdot(2x+10) -13. $$

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

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

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

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

Let's talk about the first integral first, i.e. about $\int \frac((2x+10)dx)(x^2+10x+34)$. Since $d(x^2+10x+34)=(x^2+10x+34)"dx=(2x+10)dx$, then 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. Let's single out the full square in the denominator: $x^2+10x+34=(x+5)^2+9$. In addition, we take into account $dx=d(x+5)$. Now the sum of integrals obtained by us earlier can be rewritten in a slightly different form:

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

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

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

We got the same answer as when applying the formula , which, in fact, 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, therefore I will formulate it:

Question #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:

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

Why was the module missing from the solution?

Answer to question #1

The question is completely legitimate. The modulus 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 $x^2+10x+34=(x+5)^2+9$ and $(x+5)^2 ≥ 0$, then $(x+5)^2+9 > 0$ . It is possible to judge in a different way, without involving the selection of a full square. Since $10^2-4\cdot 34=-16< 0$, 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 $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. you can use normal brackets instead of a module.

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

Answer:

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

Example #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 an elementary fraction of the third type, i.e. to $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient $3$ in front of $x^2$, but it won't take long to remove the coefficient (out of 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, so 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. And 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 the given rational fraction is not elementary, then it must be represented as a sum of elementary fractions, and then integrated. In short, trail take advantage of . How to decompose a rational fraction into elementary ones is written in detail. Let's start by factoring the denominator:

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

We represent the subinternal fraction in the following form:

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

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

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

To find the coefficients $A$ and $B$ there are two standard ways: the method of indeterminate coefficients and the method of substitution of partial values. Let's apply the partial value substitution method by substituting $x=2$ and then $x=-\frac(1)(3)$:

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

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

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

In principle, you can leave this entry, but I like a more accurate version:

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

Returning to the original integral, we substitute the resulting expansion into it. Then we divide the integral into two, and apply the formula to each. I prefer to immediately take out the constants outside the integral sign:

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

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

Example #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 is a polynomial of the second degree, and the denominator is 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:

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

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

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

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

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