## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}-40 x-8}{x(x+4)(x-2)} d x
$$
## Solution

$$
\int \frac{2 x^{3}-40 x-8}{x(x+4)(x-2)} d x=\int \frac{2 x^{3}-40 x-8}{x^{3}+2 x^{2}-8 x} d x=
$$

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

$$
\begin{array}{ll}
2 x^{3}-40 x-8 & \mid x^{3}+2 x^{2}-8 x \\
\frac{2 x^{3}+4 x^{2}-16 x}{-4 x^{2}-24 x-8} & 2
\end{array}
$$

We get:

$$
=\int\left(2+\frac{-4 x^{2}-24 x-8}{x^{3}+2 x^{2}-8 x}\right) d x=\int 2 d x+\int \frac{-4 x^{2}-24 x-8}{x(x+4)(x-2)} d x=
$$

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

$$
\begin{aligned}
& \frac{-4 x^{2}-24 x-8}{x(x+4)(x-2)}=\frac{A}{x}+\frac{B}{x+4}+\frac{C}{x-2}= \\
& =\frac{A(x+4)(x-2)+B x(x-2)+C x(x+4)}{x(x+4)(x-2)}= \\
& =\frac{A\left(x^{2}+2 x-8\right)+B\left(x^{2}-2 x\right)+C\left(x^{2}+4 x\right)}{x(x+4)(x-2)}= \\
& =\frac{(A+B+C) x^{2}+(2 A-2 B+4 C) x-8 A}{x(x+1)(x-2)}= \\
& \left\{\begin{array}{l}
A+B+C=-4 \\
2 A-2 B+4 C=-24 \\
-8 A=-8
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=-5 \\
-2 B+4 C=-26 \\
A=1
\end{array}\right.
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=-5 \\
6 C=-36 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=1 \\
C=-6 \\
A=1
\end{array}\right. \\
& \frac{-4 x^{2}-24 x-8}{x(x+4)(x-2)}=\frac{1}{x}+\frac{1}{x+4}-\frac{6}{x-2}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 2 d x+\int\left(\frac{1}{x}+\frac{1}{x+4}-\frac{6}{x-2}\right) d x= \\
& =2 x+\ln |x|+\ln |x+4|-6 \cdot \ln |x-2|+C
\end{aligned}
$$

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}+5-31 »

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

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