Division integration. Integration of rational functions Fractional - rational function The simplest
Here we provide detailed solutions of three integration examples of the following rational fractions:
,
,
.
Example 1
Calculate integral:
.
Decision
Here under the integral sign is rational function, since the integrand is a fraction of polynomials. The degree of the denominator polynomial ( 3 ) is less than the degree of the numerator polynomial ( 4 ). Therefore, first you need to select the whole part of the fraction.
1.
Let's take the integer part of the fraction. Divide x 4
on x 3 - 6 x 2 + 11 x - 6:
From here
.
2.
Let's factorize the denominator. To do this, you need to solve the cubic equation:
.
6
1, 2, 3, 6, -1, -2, -3, -6
.
Substitute x = 1
:
.
1
. Divide by x - 1
:
From here
.
We solve a quadratic equation.
.
Equation roots: , .
Then
.
3.
Let's decompose the fraction into simple ones.
.
So we found:
.
Let's integrate.
Answer
Example 2
Calculate integral:
.
Decision
Here in the numerator of the fraction is a polynomial of degree zero ( 1 = x0). The denominator is a third degree polynomial. Because the 0 < 3 , then the fraction is correct. Let's break it down into simple fractions.
1.
Let's factorize the denominator. To do this, you need to solve the equation of the third degree:
.
Assume that it has at least one integer root. Then it is the divisor of the number 3
(a member without x ). That is, the whole root can be one of the numbers:
1, 3, -1, -3
.
Substitute x = 1
:
.
So we have found one root x = 1
. Divide x 3 + 2 x - 3 on x- 1
:
So,
.
We solve the quadratic equation:
x 2 + x + 3 = 0.
Find the discriminant: D = 1 2 - 4 3 = -11. Because D< 0
, then the equation has no real roots. Thus, we have obtained the decomposition of the denominator into factors:
.
2.
.
(x - 1)(x 2 + x + 3):
(2.1)
.
Substitute x = 1
. Then x- 1 = 0
,
.
Substitute in (2.1)
x= 0
:
1 = 3 A - C;
.
Equate in (2.1)
coefficients at x 2
:
;
0=A+B;
.
.
3.
Let's integrate.
(2.2)
.
To calculate the second integral, we select the derivative of the denominator in the numerator and reduce the denominator to the sum of squares.
;
;
.
Calculate I 2
.
.
Since the equation x 2 + x + 3 = 0 has no real roots, then x 2 + x + 3 > 0. Therefore, the module sign can be omitted.
We deliver to (2.2)
:
.
Answer
Example 3
Calculate integral:
.
Decision
Here, under the sign of the integral is a fraction of polynomials. Therefore, the integrand is a rational function. The degree of the polynomial in the numerator is 3 . The degree of the polynomial of the denominator of a fraction is 4 . Because the 3 < 4 , then the fraction is correct. Therefore, it can be decomposed into simple fractions. But for this you need to decompose the denominator into factors.
1.
Let's factorize the denominator. To do this, you need to solve the equation of the fourth degree:
.
Assume that it has at least one integer root. Then it is the divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
1, 2, -1, -2
.
Substitute x = -1
:
.
So we have found one root x = -1
. Divide by x - (-1) = x + 1:
So,
.
Now we need to solve the equation of the third degree:
.
If we assume that this equation has an integer root, then it is a divisor of the number 2
(a member without x ). That is, the whole root can be one of the numbers:
1, 2, -1, -2
.
Substitute x = -1
:
.
So, we have found another root x = -1
. It would be possible, as in the previous case, to divide the polynomial by , but we will group the terms:
.
Since the equation x 2 + 2 = 0
has no real roots, then we get the factorization of the denominator:
.
2.
Let's decompose the fraction into simple ones. We are looking for a decomposition in the form:
.
We get rid of the denominator of the fraction, multiply by (x + 1) 2 (x 2 + 2):
(3.1)
.
Substitute x = -1
. Then x + 1 = 0
,
.
Differentiate (3.1)
:
;
.
Substitute x = -1
and take into account that x + 1 = 0
:
;
;
.
Substitute in (3.1)
x= 0
:
0 = 2A + 2B + D;
.
Equate in (3.1)
coefficients at x 3
:
;
1=B+C;
.
So, we have found the decomposition into simple fractions:
.
3.
Let's integrate.
.
TOPIC: Integration of rational fractions.
Attention! When studying one of the main methods of integration - the integration of rational fractions - it is required to consider polynomials in the complex domain for rigorous proofs. Therefore, it is necessary study in advance some properties of complex numbers and operations on them.
Integration of the simplest rational fractions.
If P(z) and Q(z) are polynomials in the complex domain, then is a rational fraction. It is called correct if the degree P(z) less degree Q(z) , and wrong if the degree R no less degree Q.
Any improper fraction can be represented as: ,
P(z) = Q(z) S(z) + R(z),
a R(z) – polynomial whose degree is less than the degree Q(z).
Thus, the integration of rational fractions is reduced to the integration of polynomials, that is, power functions, and proper fractions, since it is a proper fraction.
Definition 5. The simplest (or elementary) fractions are fractions of the following types:
1) , 2) , 3) , 4) .
Let's find out how they are integrated.
3) (explored earlier).
Theorem 5. Any proper fraction can be represented as a sum of simple fractions (without proof).
Corollary 1. If is a proper rational fraction, and if among the roots of the polynomial there are only simple real roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 1st type:
Example 1
Corollary 2. If is a proper rational fraction, and if among the roots of the polynomial there are only multiple real roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 1st and 2nd types:
Example 2
Corollary 3. If is a proper rational fraction, and if among the roots of the polynomial there are only simple complex conjugate roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd type:
Example 3
Corollary 4. If is a proper rational fraction, and if among the roots of the polynomial there are only multiple complex conjugate roots, then in the expansion of the fraction into the sum of simple fractions there will be only simple fractions of the 3rd and 4th types:
To determine the unknown coefficients in the above expansions, proceed as follows. The left and right parts of the expansion containing unknown coefficients are multiplied by The equality of two polynomials is obtained. Equations for the desired coefficients are obtained from it, using that:
1. equality is valid for any values of X (method of partial values). In this case, any number of equations are obtained, any m of which allow us to find unknown coefficients.
2. the coefficients coincide at the same powers of X (method of indefinite coefficients). In this case, a system of m - equations with m - unknowns is obtained, from which unknown coefficients are found.
3. combined method.
Example 5. Expand a fraction to the simplest.
Decision:
Find the coefficients A and B.
1 way - private value method:
Method 2 - the method of uncertain coefficients:
Answer:
Integration of rational fractions.
Theorem 6. The indefinite integral of any rational fraction on any interval on which its denominator is not zero, exists and is expressed in terms of elementary functions, namely rational fractions, logarithms, and arctangents.
Proof.
We represent a rational fraction in the form: . Moreover, the last term is a proper fraction, and by Theorem 5 it can be represented as a linear combination of simple fractions. Thus, integrating a rational fraction reduces to integrating a polynomial S(x) and the simplest fractions, whose antiderivatives, as was shown, have the form indicated in the theorem.
Comment. The main difficulty in this case is the decomposition of the denominator into factors, that is, the search for all its roots.
Example 1. Find the integral
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:
- $\frac(A)(x-a)$;
- $\frac(A)((x-a)^n)$ ($n=2,3,4, \ldots$);
- $\frac(Mx+N)(x^2+px+q)$ ($p^2-4q< 0$);
- $\frac(Mx+N)((x^2+px+q)^n)$ ($p^2-4q< 0$; $n=2,3,4,\ldots$).
Note (desirable for a 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 decomposition 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):
- 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, 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:
$$ 2\cdot\int \frac((2x+10)dx)(x^2+10x+34)-13\cdot\int\frac(dx)(x^2+10x+34) =2\cdot \int \frac(d(x^2+10x+34))(x^2+10x+34)-13\cdot\int\frac(d(x+5))((x+5)^2+ nine). $$
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 $\int\frac(du)(u^2+9)$. it the purest water 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:
- $\int\frac(7dx)(x+9)=7\ln|x+9|+C$;
- $\int\frac(11dx)((4x+19)^8)=-\frac(11)(28(4x+19)^7)+C$;
- $\int\frac(4x+7)(x^2+10x+34)dx=2\cdot\ln(x^2+10x+34)-\frac(13)(3)\arctg\frac(x +5)(3)+C$.
Example #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 quadratic equation$x^2+px+q=0$. If the discriminant is less than zero, then the expression $x^2+px+q$ cannot be factorized. Let's calculate the discriminant of the polynomial $3x^2-5x-2$ located in the denominator of our fraction: $D=(-5)^2-4\cdot 3\cdot(-2)=49$. So, $D > 0$, 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.
Educational Institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
Guidelines
on the study of the topic "Integration of some functions" by students of the accounting faculty absentee form education (NISPO)
Gorki, 2013
Integration of some functions
Integration of rational functions
view function
called rational fraction
, if its numerator and denominator are polynomials. rational fraction called correct
if the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, then the rational fraction called wrong
.
Since any improper fraction can be represented as the sum of a polynomial and a proper fraction, then integrating an improper rational fraction is reduced to integrating a polynomial and a proper rational fraction.
Polynomials are easy to integrate. Consider the integration of fractions of the form
,
, which are called the simplest rational fractions
.
.
.
Let the denominator
fraction has real roots and can be represented by a product of factors of the form
. Then for each such factor there is a decomposition of the form
. Thus, every proper rational fraction can be represented as the sum of a finite number of simple fractions. This is done using the method of undetermined coefficients.
Example 1
. Integrate a fraction
.
Decision
.
We expand the integrand into simple fractions:
Equate the coefficients at and free members:
We solve this system of equations and get ,
. Then
.
Integration of some irrational functions
If the integrand is irrational, then by changing the variable, in many cases it can be reduced to a rational form or to a function whose integral is tabular. Integration by means of a change of variable that reduces the integrand to a rational form is called integration via rationalization of the integrand .
Integrals of the form
are reduced to integrals of rational functions of the argument t using substitution
, where k is the least common multiple of the numbers
.
Example 2
. Find the integral
.
Decision
. Least common multiple of numbers
and
equals 6. Therefore, you need to apply the substitution
. Then
. We decompose the integrand into the simplest ones: . Equate the coefficients at and free members:
From here we find
Then
. So =
. As
, then
. Substitute in the resulting expression:
.
Integrals of the form
are reduced to integrals of rational functions using the substitution
.
Example 3
. Find the integral
.
Decision
. Let's perform the substitution
:
.
Integration of expressions containing
trigonometric functions
Consider the main cases of integration of expressions containing trigonometric functions.
When finding integrals of the form
,
,
integrands from prod-
leads are converted to amounts using the formulas:
As a result, the resulting integrals are found using integration methods and the table of integrals. In this case, you can use the formulas
and
.
Example 4
. Find the integral
.
Decision . Let's use the first of the above formulas:
Integrals of the form
can be found quite simply in the following cases.
If m is a positive odd number, then you can separate the first power of the sine and apply the substitution
. Then
and the integrand with the help of trigonometric formulas will be reduced to power functions. If n is a positive odd number, then you can separate the first power of the cosine and perform the replacement
. Then
and the integrand with the help of trigonometric functions also reduces to power functions.
Example 5
. Find the integral
.
Decision .
.
Example 6
. Find the integral
.
Decision .
If m and n are non-negative even numbers, then the transformation of the integrands can be performed using the power reduction formulas
and
.
Example 7
. Find the integral
.
Decision .
.
The integrand is a fraction whose numerator is the degree of the sine and the denominator is the degree of the cosine, or vice versa. In this case, the exponents are either both even or both odd, i.e. the same parity.
In this case, if the numerator is a sine, then the most appropriate substitution is
. From here
,
,
,
.
If the numerator has a cosine, then it is convenient to use the substitution
. Then
,
,
,
.
Example 8
. Find the integral
.
Decision .
.
Finding integrals of the form
reduced by substitution
to finding integrals of rational functions. Substitution
called universal trigonometric substitution
, which always leads to a result. In this case
,
,
,
,
.
Example 9
. Find the integral
.
Decision
.
.
Questions for self-control of knowledge
are reduced to integrals of rational functions?
,
What is the universal trigonometric substitution and when is it used?
Tasks for independent work
Find integrals of rational functions:
a)
; b)
; in)
.
2) Integrate expressions containing trigonometric functions:
a)
; b)
; in)
;
G)
; e)
.
2.,
5.
,
3.
,
6.
.
In integrals 1-3 as u accept . Then, after n-fold application of formula (19), we arrive at one of the table integrals
,
,
.
In integrals 4-6, when differentiating, the transcendental factor is simplified
,
or
, which should be taken as u.
Calculate the following integrals.
Example 7
Example 8
Reducing integrals to itself
If the integrand
looks like:
,
,
and so on,
then after double integration by parts we obtain an expression containing the original integral :
,
where
is some constant.
Solving the resulting equation with respect to , we obtain a formula for calculating the original integral:
.
This case of applying the method of integration by parts is called " bringing the integral to itself».
Example 9 Calculate Integral
.
On the right side is the original integral . Moving it to the left side, we get:
.
Example 10 Calculate Integral
.
4.5. Integration of the simplest proper rational fractions
Definition.The simplest proper fractions I , II and III types the following fractions are called:
I. ;
II.
;
(
is a positive integer);
III.
; (the roots of the denominator are complex, that is:
.
Consider integrals of simple fractions.
I.
;
(20)
II. ; (21)
III.
;
We transform the numerator of the fraction in such a way as to single out the term in the numerator
equal to the derivative of the denominator.
Consider the first of the two integrals obtained and make a change in it:
In the second integral, we complement the denominator to a full square:
Finally, the integral of a fraction of the third type is equal to:
=
+
.
(22)
Thus, the integral of the simplest fractions of type I is expressed in terms of logarithms, type II - in terms of rational functions, type III - in terms of logarithms and arctangents.
4.6 Integration of fractional-rational functions
One of the classes of functions that have an integral expressed in terms of elementary functions is the class of algebraic rational functions, that is, functions resulting from a finite number of algebraic operations on an argument.
Every rational function
can be represented as a ratio of two polynomials
and
:
. (23)
We will assume that the polynomials do not have common roots.
A fraction of the form (23) is called correct, if the degree of the numerator is less than the degree of the denominator, that is, m< n. Otherwise - wrong.
If the fraction is incorrect, then, dividing the numerator by the denominator (according to the rule of dividing polynomials), we represent the fraction as the sum of a polynomial and a proper fraction:
, (24)
where
- polynomial, is a proper fraction, and the degree of the polynomial
- no higher degree ( n-1).
Example.
Since the integration of a polynomial is reduced to the sum of table integrals from power function, then the main difficulty in integrating rational fractions is integrating proper rational fractions.
Algebra proves that every proper fraction decomposes into the sum of the above protozoa fractions, the form of which is determined by the roots of the denominator
.
Let's consider three special cases. Here and below, we will assume that the coefficient at the highest degree of the denominator
equal to one =1, that isreduced polynomial
.
Case 1 The roots of the denominator, that is, the roots
equations
=0 are real and different. Then we represent the denominator as a product of linear factors:
and the proper fraction decomposes into the simplest fractions of the I-type:
, (26)
where
are some constant numbers that are found by the method of indefinite coefficients.
For this you need:
1. Reduce the right side of expansion (26) to a common denominator.
2. Equate the coefficients at the same powers of the identical polynomials in the numerator of the left and right parts. We obtain a system of linear equations for determining
.
3. Solve the resulting system and find the uncertain coefficients
.
Then the integral of the fractional-rational function (26) will be equal to the sum of the integrals of the simplest fractions of the I-type, calculated by formula (20).
Example. Calculate Integral
.
Decision. Let's factorize the denominator using Vieta's theorem:
Then, the integrand expands into the sum of simple fractions:
.
X:
Let us write a system of three equations for finding
X on the left and right sides:
.
Let us indicate a simpler method for finding indeterminate coefficients, called partial value method.
Assuming in equality (27)
we get
, where
. Assuming
we get
. Finally, assuming
we get
.
.
Case 2 denominator root
are real, but among them there are multiple (equal) roots. Then we represent the denominator as a product of linear factors included in the product to the extent that the multiplicity of the corresponding root is:
where
.
Proper fraction the sum of fractions of the I-th and II-th types will be expanded. Let, for example, - root of the multiplicity denominator k, and all the rest ( n- k) of roots are different.
Then the decomposition will look like:
Similarly, if there are other multiple roots. For non-multiple roots, the expansion (28) includes the simplest fractions of the first type.
Example. Calculate Integral
.
Decision. Let's represent a fraction as a sum of simple fractions of the first and second kind with indefinite coefficients:
.
We bring the right side to a common denominator and equate the polynomials in the numerators of the left and right sides:
On the right-hand side, we give similar ones with the same degrees X:
Let us write down the system of four equations for finding
and . To do this, we equate the coefficients at the same powers X on the left and right side
.
Case 3 Among the roots of the denominator
have complex one-time roots. That is, the expansion of the denominator includes factors of the second degree
, which cannot be decomposed into real linear factors, and they do not repeat.
Then, in the expansion of the fraction, each such factor will correspond to the simplest type III fraction. Linear factors correspond to the simplest fractions of the I-th and II-th types.
Example. Calculate Integral
.
Decision.
.
.
.
- "Fontanka" learned about the problems with weapons and salaries of PMC Wagner in Syria Islamic Anti-Terrorist Coalition
- Husband of ZSRO, honored lawyer Irina Rukavishnikova, businessman Andrey Korovaiko fled to Spain Korovaiko's wedding in Spain
- Kapinus Nikolai what are the main tasks of Nostra for the near future
- Sergunin vs. Sergunina Housing and communal services and transport