The fraction is called correct, if the highest degree of the numerator is less than the highest degree 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 extract a complete 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 the following 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 form by dividing the polynomial from the numerator by the polynomial from the denominator. IN in this case the formula for integrating a rational fraction has the form:
$$ \int \frac(P(x))(ax^2+bx+c)dx = \int Q(x) dx + \int \frac(mx+n)(ax^2+bx+c)dx $$
Examples of solutions
| Example 1 |
| Find the integral of the rational fraction: $$ \int \frac(dx)(x^2-10x+16) $$ |
| Solution |
|
The fraction is proper and the polynomial has only complex roots. Therefore, we select a complete square: $$ \int \frac(dx)(x^2-10x+16) = \int \frac(dx)(x^2-2\cdot 5 x+ 5^2 - 9) = $$ We fold a complete square and place it under the differential sign $ x-5 $: $$ = \int \frac(dx)((x-5)^2 - 9) = \int \frac(d(x-5))((x-5)^2-9) = $$ Using the table of integrals we obtain: $$ = \frac(1)(2 \cdot 3) \ln \bigg | \frac(x-5 - 3)(x-5 + 3) \bigg | + C = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ \int \frac(dx)(x^2-10x+16) = \frac(1)(6) \ln \bigg |\frac(x-8)(x-2) \bigg | +C$$ |
| Example 2 |
| Perform integration of rational fractions: $$ \int \frac(x+2)(x^2+5x-6) dx $$ |
| Solution |
|
Let's solve the quadratic equation: $$ x^2+5x-6 = 0 $$ $$ x_(12) = \frac(-5\pm \sqrt(25-4\cdot 1 \cdot (-6)))(2) = \frac(-5 \pm 7)(2) $$ We write down the roots: $$ x_1 = \frac(-5-7)(2) = -6; x_2 = \frac(-5+7)(2) = 1 $$ Taking into account the obtained roots, we transform the integral: $$ \int \frac(x+2)(x^2+5x-6) dx = \int \frac(x+2)((x-1)(x+6)) dx = $$ We perform the expansion of a rational fraction: $$ \frac(x+2)((x-1)(x+6)) = \frac(A)(x-1) + \frac(B)(x+6) = \frac(A(x -6)+B(x-1))((x-1)(x+6)) $$ We equate the numerators and find the coefficients $ A $ and $ B $: $$ A(x+6)+B(x-1)=x+2 $$ $$ Ax + 6A + Bx - B = x + 2 $$ $$ \begin(cases) A + B = 1 \\ 6A - B = 2 \end(cases) $$ $$ \begin(cases) A = \frac(3)(7) \\ B = \frac(4)(7) \end(cases) $$ We substitute the found coefficients into the integral and solve it: $$ \int \frac(x+2)((x-1)(x+6))dx = \int \frac(\frac(3)(7))(x-1) dx + \int \frac (\frac(4)(7))(x+6) dx = $$ $$ = \frac(3)(7) \int \frac(dx)(x-1) + \frac(4)(7) \int \frac(dx)(x+6) = \frac(3) (7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$ |
| Answer |
| $$ \int \frac(x+2)(x^2+5x-6) dx = \frac(3)(7) \ln |x-1| + \frac(4)(7) \ln |x+6| +C$$ |
Let us remind you that fractional-rational are called functions of the form $$ f(x) = \frac(P_n(x))(Q_m(x)), $$ in general case which are the ratio of two polynomials %%P_n(x)%% and %%Q_m(x)%%.
If %%m > n \geq 0%%, then the rational fraction is called correct, otherwise - incorrect. Using the rule for dividing polynomials, an improper rational fraction can be represented as the sum of a polynomial %%P_(n - m)%% of degree %%n - m%% and some proper fraction, i.e. $$ \frac(P_n(x))(Q_m(x)) = P_(n-m)(x) + \frac(P_l(x))(Q_n(x)), $$ where the degree %%l%% of the polynomial %%P_l(x)%% is less than the degree %%n%% of the polynomial %%Q_n(x)%%.
Thus, the indefinite integral of a rational function can be represented as the sum of the indefinite integrals of a polynomial and a proper rational fraction.
Integrals from simple rational fractions
Among proper rational fractions, there are four types, which are classified as simple rational fractions:
- %%\displaystyle \frac(A)(x - a)%%,
- %%\displaystyle \frac(A)((x - a)^k)%%,
- %%\displaystyle \frac(Ax + B)(x^2 + px + q)%%,
- %%\displaystyle \frac(Ax + B)((x^2 + px + q)^k)%%,
where %%k > 1%% is an integer and %%p^2 - 4q< 0%%, т.е. квадратные уравнения не имеют действительных корней.
Calculation of indefinite integrals of fractions of the first two types
Calculating indefinite integrals of fractions of the first two types does not cause difficulties: $$ \begin(array)(ll) \int \frac(A)(x - a) \mathrm(d)x &= A\int \frac(\mathrm (d)(x - a))(x - a) = A \ln |x - a| + C, \\ \\ \int \frac(A)((x - a)^k) \mathrm(d)x &= A\int \frac(\mathrm(d)(x - a))(( x - a)^k) = A \frac((x-a)^(-k + 1))(-k + 1) + C = \\ &= -\frac(A)((k-1)(x-a )^(k-1)) + C. \end(array) $$
Calculation of indefinite integrals of fractions of the third type
We first transform the third type of fraction by highlighting the perfect square in the denominator: $$ \frac(Ax + B)(x^2 + px + q) = \frac(Ax + B)((x + p/2)^2 + q - p^2/4), $$ since %%p^2 - 4q< 0%%, то %%q - p^2/4 >0%%, which we denote as %%a^2%%. Having also replaced %%t = x + p/2, \mathrm(d)t = \mathrm(d)x%%, we transform the denominator and write the integral of the third type fraction in the form $$ \begin(array)(ll) \ int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x &= \int \frac(Ax + B)((x + p/2)^2 + q - p^2 /4) \mathrm(d)x = \\ &= \int \frac(A(t - p/2) + B)(t^2 + a^2) \mathrm(d)t = \int \frac (At + (B - A p/2))(t^2 + a^2) \mathrm(d)t. \end(array) $$
Using the linearity of the indefinite integral, we represent the last integral as a sum of two and in the first of them we introduce %%t%% under the differential sign: $$ \begin(array)(ll) \int \frac(At + (B - A p /2))(t^2 + a^2) \mathrm(d)t &= A\int \frac(t \mathrm(d)t)(t^2 + a^2) + \left(B - \frac(pA)(2)\right)\int \frac(\mathrm(d)t)(t^2 + a^2) = \\ &= \frac(A)(2) \int \frac( \mathrm(d)\left(t^2 + a^2\right))(t^2 + a^2) + - \frac(2B - pA)(2)\int \frac(\mathrm(d) t)(t^2 + a^2) = \\ &= \frac(A)(2) \ln \left| t^2 + a^2\right| + \frac(2B - pA)(2a) \text(arctg)\frac(t)(a) + C. \end(array) $$
Returning to the original variable %%x%%, as a result, for a fraction of the third type we obtain $$ \int \frac(Ax + B)(x^2 + px + q) \mathrm(d)x = \frac(A)( 2) \ln \left| x^2 + px + q\right| + \frac(2B - pA)(2a) \text(arctg)\frac(x + p/2)(a) + C, $$ where %%a^2 = q - p^2 / 4 > 0% %.
Calculating a type 4 integral is difficult and is therefore not covered in this course.
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 highly recommend that you at least skim through this topic before moving on to reading. of 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 corresponding 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 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 four types:
- $\frac(A)(x-a)$;
- $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
- $\frac(Mx+N)(x^2+px+q)$ ($p^2-4q< 0$);
- $\frac(Mx+N)((x^2+px+q)^n)$ ($p^2-4q< 0$; $n=2,3,4,\ldots$).
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: $p^2-4q=5^2-4\cdot 10=-15$. 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 at all necessary that the coefficient before $x^2$ be equal to 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 (proper and improper), as well as examples of decomposition of a rational fraction into elementary ones can be found. Here we will be interested only in questions of their integration. Let's start with the integration of elementary fractions. So, each of the four types of elementary fractions above is easy to integrate using the formulas below. Let me remind you that when integrating fractions of types (2) and (4), $n=2,3,4,\ldots$ are assumed. Formulas (3) and (4) require the fulfillment of 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 substitution $t=x+\frac(p)(2)$ is made, after which the resulting interval is divided into two. The first will be calculated by entering under the differential sign, and the second will have the form $I_n=\int\frac(dt)((t^2+a^2)^n)$. 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 discussed in example No. 7 (see the third part).
Scheme for calculating integrals of rational functions (rational fractions):
- If the integrand is elementary, then apply formulas (1)-(4).
- 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 you can integrate any rational fraction. That is why almost all changes of variables in the indefinite integral (Euler, Chebyshev substitutions, universal trigonometric substitution) are made in such a way that after this replacement we obtain a rational fraction under the interval. And then 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 $dx=d(x+9)$, 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 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 it “manually”.
2) Again, there are two ways: use the ready-made formula or do without it. If you apply the formula, you should take into account that the coefficient in front of $x$ (number 4) will have to be removed. To do this, let’s simply take this four out of 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 taking the constant $4$ out of brackets. If we take into account that $dx=\frac(1)(4)d(4x+19)$, 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 for finding such integrals are given in the topic “Integration by substitution (substitution 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 really an elementary fraction of the third type, you need to check that the condition $p^2-4q is met< 0$. 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 a 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 will not last long. Let's 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 $2x+10$ appears 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 split the integrand into two. Well, and, accordingly, the integral itself is also “bifurcated”:
$$ \int\frac(2\cdot(2x+10)-13)(x^2+10x+34)dx=\int \left(\frac(2\cdot(2x+10))(x^2 +10x+34)-\frac(13)(x^2+10x+34) \right)\; dx=\\ =\int \frac(2\cdot(2x+10))(x^2+10x+34)dx-\int\frac(13dx)(x^2+10x+34)=2\cdot \int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34). $$
Let's first talk about the first integral, 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 numerator of the integrand contains the differential of the denominator. 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 select a complete 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 we obtained earlier can be rewritten in a slightly different form:
$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ 9). $$
If we make the substitution $u=x^2+10x+34$ in the first integral, then it will take the form $\int\frac(du)(u)$ and take easy to use second formula from . As for the second integral, the change $u=x+5$ is feasible for it, after which it will take the form $\int\frac(du)(u^2+9)$. This pure water eleventh formula from the table of indefinite integrals. So, returning to the sum of integrals, we 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 received 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 the attentive reader may have one question here, so I will formulate it:
Question No. 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 there no module in the solution?
Answer to question #1
The question is completely natural. The module was missing 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 $(x+5)^2+9 > 0$ . You can think differently, without using the selection of a complete 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 quadratic inequalities). In any case, since $x^2+10x+34 > 0$, then $|x^2+10x+34|=x^2+10x+34$, i.e. Instead of a module, you can use regular brackets.
All points of example No. 1 have been solved, all that remains is to write down the answer.
Answer:
- $\int\frac(7dx)(x+9)=7\ln|x+9|+C$;
- $\int\frac(11dx)((4x+19)^8)=-\frac(11)(28(4x+19)^7)+C$;
- $\int\frac(4x+7)(x^2+10x+34)dx=2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x +5)(3)+C$.
Example No. 2
Find the integral $\int\frac(7x+12)(3x^2-5x-2)dx$.
At first glance, the integrand fraction $\frac(7x+12)(3x^2-5x-2)$ is very similar to an elementary fraction of the third type, i.e. by $\frac(Mx+N)(x^2+px+q)$. It seems that the only difference is the coefficient of $3$ in front of $x^2$, but it doesn’t take long to remove the coefficient (put it out of brackets). However, this similarity is apparent. For the fraction $\frac(Mx+N)(x^2+px+q)$ the condition $p^2-4q is mandatory< 0$, 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 before $x^2$ 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, $D >0$, therefore the expression $3x^2-5x-2$ can be factorized. This means that the fraction $\frac(7x+12)(3x^2-5x-2)$ is not an elemental fraction of the third type, and apply $\int\frac(7x+12)(3x^2-) to the integral 5x-2)dx$ formula is not possible.
Well, if the given rational fraction is not an elementary fraction, then it needs to 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:
$$ 3x^2-5x-2=0;\\ \begin(aligned) & D=(-5)^2-4\cdot 3\cdot(-2)=49;\\ & x_1=\frac( -(-5)-\sqrt(49))(2\cdot 3)=\frac(5-7)(6)=\frac(-2)(6)=-\frac(1)(3); \\ & x_2=\frac(-(-5)+\sqrt(49))(2\cdot 3)=\frac(5+7)(6)=\frac(12)(6)=2.\ \\end(aligned)\\ 3x^2-5x-2=3\cdot\left(x-\left(-\frac(1)(3)\right)\right)\cdot (x-2)= 3\cdot\left(x+\frac(1)(3)\right)(x-2). $$
We present the subintercal fraction in this 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 decompose 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 undetermined coefficients and the method of substitution of partial values. Let's apply the partial value substitution method, 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, all that remains is 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 option:
$$ \frac(\frac(7)(3)x+4)(\left(x+\frac(1)(3)\right)(x-2))=-\frac(29)(21)\ cdot\frac(1)(x+\frac(1)(3))+\frac(26)(7)\cdot\frac(1)(x-2). $$
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 place 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 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:
$$ \frac(x^2-38x+157)((x-1)(x+4)(x-9))=-\frac(3)(x-1)+\frac(5)(x +4)-\frac(1)(x-9). $$
All we have to do is split the given integral into three and apply the formula to each. I prefer to immediately place 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: $\int\frac(x^2-38x+157)((x-1)(x+4)(x-9))dx=-3\ln|x-1|+5\ln|x+ 4|-\ln|x-9|+C$.
Continuation of the analysis of examples of this topic is located in the second part.
To integrate a rational function \(\large\frac((P\left(x \right)))((Q\left(x \right)))\normalsize,\) where \((P\left(x \right ))\) and \((Q\left(x \right))\) are polynomials, the following sequence of steps is used:
If the fraction is an improper fraction (i.e. the degree of \((P\left(x \right))\) is greater than the degree of \((Q\left(x \right))\)), convert it to a proper fraction by highlighting the whole expression;
Expand the denominator \((Q\left(x \right))\) into the product of monomials and/or irreducible quadratic expressions;
Resolve a rational fraction into simpler fractions using ;
Calculate integrals of simple fractions.
Step 1: Converting an improper rational fraction
If the fraction is improper (i.e. the degree of the numerator \((P\left(x \right))\) is greater than the degree of the denominator \((Q\left(x \right))\)), divide the polynomial \((P\ left(x \right))\) by \((Q\left(x \right)).\) We obtain the following expression: \[\frac((P\left(x \right)))((Q\left (x \right))) = F\left(x \right) + \frac((R\left(x \right)))((Q\left(x \right))),\] where \(\ large\frac((R\left(x \right)))((Q\left(x \right)))\normalsize\) is a proper rational fraction.
Step 2. Decomposition of the denominator into simple fractions
Let us write the denominator polynomial \((Q\left(x \right))\) in the form \[ (Q\left(x \right) ) = ((\left((x - a) \right)^\alpha ) \ cdots (\left((x - b) \right)^\beta )(\left(((x^2) + px + q) \right)^\mu ) \cdots (\left(((x^2 ) + rx + s) \right)^\nu ),) \] where quadratic functions are irreducible, that is, having no real roots.
Step 3. Decomposition of a rational fraction into a sum of simple fractions.
Let us write the rational function in the following form: \[ (\frac((R\left(x \right)))((Q\left(x \right))) = \frac(A)((((\left(( x - a) \right))^\alpha ))) + \frac(((A_1)))((((\left((x - a) \right))^(\alpha - 1)))) + \ldots )\kern0pt (+ \frac(((A_(\alpha - 1))))((x - a)) + \ldots )\kern0pt (+ \frac(B)((((\left( (x - b) \right))^\beta ))) + \frac(((B_1)))((((\left((x - b) \right))^(\beta - 1))) ) + \ldots )\kern0pt (+ \frac(((B_(\beta - 1))))((x - b)) )\kern0pt (+ \frac((Kx + L))((((\ left(((x^2) + px + q) \right))^\mu ))) + \frac(((K_1)x + (L_1)))((((\left(((x^2 ) + px + q) \right))^(\mu - 1)))) + \ldots )\kern0pt (+ \frac(((K_(\mu - 1))x + (L_(\mu - 1 ))))(((x^2) + px + q)) + \ldots )\kern0pt (+ \frac((Mx + N))((((\left(((x^2) + rx + s) \right))^\nu ))) + \frac(((M_1)x + (N_1)))((((\left(((x^2) + rx + s) \right))^ (\nu - 1)))) + \ldots )\kern0pt (+ \frac(((M_(\nu - 1))x + (N_(\nu - 1))))(((x^2) + rx + s)).) \] Total number uncertain coefficients\((A_i),\) \((B_i),\) \((K_i),\) \((L_i),\) \((M_i),\) \((N_i), \ldots\) must be equal to the degree of the denominator \((Q\left(x \right)).\)
Then we multiply both sides of the resulting equation by the denominator \((Q\left(x \right))\) and equate the coefficients of terms with the same powers \(x.\) As a result, we obtain the system linear equations with respect to unknown coefficients \((A_i),\) \((B_i),\) \((K_i),\) \((L_i),\) \((M_i),\) \((N_i), \ ldots\) This system always has the only solution. The described algorithm is method of uncertain coefficients .
Step 4. Integration of simple rational fractions.
The simplest fractions obtained by decomposing an arbitrary proper rational fraction are integrated using the following six formulas: \ \ For fractions with a quadratic denominator, you first need to isolate the perfect square: \[\int (\frac((Ax + B))((((\ left(((x^2) + px + q) \right))^k)))dx) = \int (\frac((At + B"))((((\left(((t^2 ) + (m^2)) \right))^k)))dt) ,\] where \(t = x + \large\frac(p)(2)\normalsize,\) \((m^2 ) = \large\frac((4q - (p^2)))(4)\normalsize,\) \(B" = B - \large\frac((Ap))(2)\normalsize.\) Then the following formulas are used: \ \[ (4.\;\;\int (\frac((tdt))((((\left(((t^2) + (m^2)) \right))^k )))) ) = (\frac(1)((2\left((1 - k) \right)((\left(((t^2) + (m^2)) \right))^( k - 1)))) ) \] \Integral \(\large\int\normalsize (\large\frac((dt))((((\left(((t^2) + (m^2)) \right))^k)))\normalsize) \) can be calculated in \(k\) steps using reduction formulas\[ (6.\;\;\int (\frac((dt))((((\left(((t^2) + (m^2)) \right))^k)))) ) = (\frac(t)((2(m^2)\left((k - 1) \right)((\left(((t^2) + (m^2)) \right))^( k - 1)))) ) (+ \frac((2k - 3))((2(m^2)\left((k - 1) \right)))\int (\frac((dt)) ((((\left(((t^2) + (m^2)) \right))^(k - 1))))) ) \]
“A mathematician, just like an artist or poet, creates patterns. And if his patterns are more stable, it is only because they are composed of ideas... The patterns of a mathematician, just like the patterns of an artist or a poet, must be beautiful; Ideas, just like colors or words, must correspond to each other. Beauty is the first requirement: there is no place in the world for ugly mathematics».
G.H.Hardy
In the first chapter it was noted that there are antiderivatives of fairly simple functions that can no longer be expressed through elementary functions. In this regard, those classes of functions about which we can accurately say that their antiderivatives are elementary functions acquire enormous practical importance. This class of functions includes rational functions, representing the ratio of two algebraic polynomials. Many problems 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) is called the relation of two algebraic polynomials:
where and are polynomials.
Let us remind you that polynomial (polynomial, whole rational function) nth degree called a function of the form
Where 
– real numbers. For example,
– polynomial of the first degree;
– polynomial of the fourth degree, etc.
The rational fraction (2.1.1) is called correct, if the degree is lower than the degree , i.e. n<m, otherwise the fraction is called wrong.
Any improper fraction can be represented as the sum of a polynomial (integer part) and a proper fraction (fractional part). The separation of the whole and fractional parts of an improper fraction can be done according to the rule for dividing polynomials with a “corner”.
Example 2.1.1. Identify the whole and fractional parts of the following improper rational fractions:
A) 
, b) 
.
Solution . a) Using the “corner” division algorithm, we get
Thus, we get
.
b) Here we also use the “corner” division algorithm:
As a result, we get
.
Let's summarize. In the general case, the indefinite integral of a rational fraction can be represented as the sum of the integrals of the polynomial and the proper rational fraction. Finding antiderivatives of polynomials is not difficult. Therefore, in what follows we will mainly consider proper rational fractions.
2.1.2. The simplest rational fractions and their integration
Among proper rational fractions, there are four types, which are classified as the simplest (elementary) rational fractions:
|
3) |
4) |
,
,where is an integer, 
, i.e. quadratic trinomial 
has no real roots.
Integrating simple fractions of the 1st and 2nd types does not present any great difficulties:
, (2.1.3)
. (2.1.4)
Let us now consider the integration of simple fractions of the 3rd type, but we will not consider fractions of the 4th type.
Let's start with integrals of the form
.
This integral is usually calculated by isolating the perfect square of the denominator. The result is a table integral of the following form
or 
.
Example 2.1.2. Find the integrals:
A) 
, b) 
.
Solution . a) Select a complete square from a quadratic trinomial:
From here we find
b) By isolating a complete square from a quadratic trinomial, we obtain:
Thus,
.
To find the integral
you can isolate the derivative of the denominator in the numerator and expand the integral into the sum of two integrals: the first of them by substitution 
comes down to appearance
,
and the second - to the one discussed above.
Example 2.1.3. Find the integrals:
.
Solution
. Note that 
. Let us isolate the derivative of the denominator in the numerator:
The first integral is calculated using substitution 
:
In the second integral, we select the perfect square in the denominator
Finally, we get
2.1.3. Proper rational fraction expansion
for the sum of simple fractions
Any proper rational fraction 
can be represented in a unique way as a sum of simple fractions. To do this, the denominator must be factorized. From higher algebra it is known that every polynomial with real coefficients
