## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-2) x} d x
$$
## Solution

$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-2) x} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-6 x^{2}+8 x} d x=
$$

Under the integral, we have an improper fraction. Let's separate the integer part:

$$
\begin{array}{ll}
x^{3}-3 x^{2}-12 & \mid x^{3}-6 x^{2}+8 x \\
\frac{x^{3}-6 x^{2}+8 x}{3 x^{2}-8 x}-12
\end{array}
$$

We get:

$$
=\int\left(1+\frac{3 x^{2}-8 x-12}{x^{3}-6 x^{2}+8 x}\right) d x=\int d x+\int \frac{3 x^{2}-8 x-12}{(x-4)(x-2) x} d x=
$$

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

$$
\begin{aligned}
& \frac{3 x^{2}-8 x-12}{(x-4)(x-2) x}=\frac{A}{x-4}+\frac{B}{x-2}+\frac{C}{x}= \\
& =\frac{A(x-2) x+B(x-4) x+C(x-4)(x-2)}{(x-4)(x-2) x}= \\
& =\frac{A\left(x^{2}-2 x\right)+B\left(x^{2}-4 x\right)+C\left(x^{2}-6 x+8\right)}{(x-4)(x-2) x}= \\
& =\frac{(A+B+C) x^{2}+(-2 A-4 B-6 C) x+8 C}{(x-4)(x-2) x}= \\
& \left\{\begin{array}{l}
A+B+C=3 \\
-2 A-4 B-6 C=-8 \\
8 C=-12
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
A+B=\frac{9}{2} \\
-2 A-4 B=-17 \\
C=-\frac{3}{2}
\end{array}\right.
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B=\frac{9}{2} \\
2 A=1 \\
C=-\frac{3}{2}
\end{array}\right. \\
& \left\{\begin{array}{l}
B=4 \\
A=\frac{1}{2} \\
C=-\frac{3}{2}
\end{array}\right. \\
& \frac{3 x^{2}-8 x+12}{(x-4)(x-2) x}=\frac{1}{2(x-4)}+\frac{4}{x-2}-\frac{3}{2 x}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int d x+\int\left(\frac{1}{2(x-4)}+\frac{4}{x-2}-\frac{3}{2 x}\right) d x= \\
& =x+\frac{1}{2} \cdot \ln |x-4|+4 \cdot \ln |x-2|-\frac{3}{2} \cdot \ln |x|+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-14$ »

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

- Last modified: 09:21, 6 April 2009.
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