## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x(x-2)(x+2)} d x
$$
## Solution

$$
\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x(x-2)(x+2)} d x=\int \frac{2 x^{4}-5 x^{2}-8 x-8}{x^{3}-4 x} d x=
$$

The integrand is an improper fraction. We will separate the integer part:

$$
\begin{array}{ll}
2 x^{4}-5 x^{2}-8 x-8 & \left\lvert\, \frac{x^{3}-4 x}{2 x}\right. \\
\frac{2 x^{4}-8 x^{2}}{3 x^{2}-8 x-8} &
\end{array}
$$

We get:

$$
=\int\left(2 x+\frac{3 x^{2}-8 x-8}{x^{3}-4 x}\right) d x=\int 2 x d x+\int \frac{3 x^{2}-8 x-8}{x(x-2)(x+2)} d x=
$$

We will decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{3 x^{2}-8 x-8}{x(x-2)(x+2)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2}= \\
& =\frac{A(x-2)(x+2)+B x(x+2)+C x(x-2)}{x(x-2)(x+2)}= \\
& =\frac{A\left(x^{2}-4\right)+B\left(x^{2}+2 x\right)+C\left(x^{2}-2 x\right)}{x(x-2)(x+2)}= \\
& =\frac{(A+B+C) x^{2}+(2 B-2 C) x-4 A}{x(x-2)(x+2)}= \\
& \left\{\begin{array}{l}
A+B+C=3 \\
2 B-2 C=-8 \\
-4 A=-8
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=1 \\
2 B-2 C=-8 \\
A=2
\end{array}\right.
$$

Add the first row multiplied by 2 to the second row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=1 \\
4 B=-6 \\
A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
C=\frac{5}{2} \\
B=-\frac{3}{2} \\
A=2
\end{array}\right. \\
& \frac{3 x^{2}-8 x-8}{x(x-2)(x+2)}=\frac{2}{x}-\frac{3}{2(x-2)}+\frac{5}{2(x+2)}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 2 x d x+\int\left(\frac{2}{x}-\frac{3}{2(x-2)}+\frac{5}{2(x+2)}\right) d x= \\
& =x^{2}+2 \cdot \ln |x|-\frac{3}{2} \cdot \ln |x-2|+\frac{5}{2} \cdot \ln |x+2|+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-23$ "

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

- Last modified: 20:42, 6 April 2009.
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