## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x(x-4)(x+5)} d x
$$
## Solution

$$
\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x(x-4)(x+5)} d x=\int \frac{2 x^{4}+2 x^{3}-41 x^{2}+20}{x^{3}+x^{2}-20 x} d x=
$$

The integrand is an improper fraction. Let's separate the integer part:

$$
\begin{array}{ll}
2 x^{4}+2 x^{3}-41 x^{2}+20 & \mid x^{3}+x^{2}-20 x \\
\frac{2 x^{4}+2 x^{3}-40 x^{2}}{2 x} & -x^{2}+20
\end{array}
$$

We get:

$$
=\int\left(2 x+\frac{-x^{2}+20}{x^{3}+x^{2}-20 x}\right) d x=\int 2 x d x+\int \frac{-x^{2}+20}{x(x-4)(x+5)} d x=
$$

Let's decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{-x^{2}+20}{x(x-4)(x+5)}=\frac{A}{x}+\frac{B}{x-4}+\frac{C}{x+5}= \\
& =\frac{A(x-4)(x+5)+B x(x+5)+C x(x-4)}{x(x-4)(x+5)}= \\
& =\frac{A\left(x^{2}+x-20\right)+B\left(x^{2}+5 x\right)+C\left(x^{2}-4 x\right)}{x(x-4)(x+5)}= \\
& =\frac{(A+B+C) x^{2}+(A+5 B-4 C) x-20 A}{x(x-4)(x+5)}= \\
& \left\{\begin{array}{l}
A+B+C=-1 \\
A+5 B-4 C=0 \\
-20 A=20
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=0 \\
5 B-4 C=1 \\
A=-1
\end{array}\right.
$$

Add to the second row the first row multiplied by 4:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=0 \\
9 B=1 \\
A=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-\frac{1}{9} \\
B=\frac{1}{9} \\
A=-1
\end{array}\right. \\
& \frac{-x^{2}+20}{x(x-4)(x+5)}=-\frac{1}{x}+\frac{1}{9(x-4)}-\frac{1}{9(x+5)}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 2 x d x+\int\left(-\frac{1}{x}+\frac{1}{9(x-4)}-\frac{1}{9(x+5)}\right) d x= \\
& =x^{2}-\ln |x|+\frac{1}{9} \cdot \ln |x-4|-\frac{1}{9} \cdot \ln |x+5|+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-26 \%$

Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

- Last modified: 21:20, 6 April 2009.
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## Problem Kuznetsov Integrals 5-27

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