## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}}{(x-1)(x+1)(x+2)} d x
$$
## Solution

$$
\begin{aligned}
& \int \frac{x^{3}}{(x-1)(x+1)(x+2)} d x=\int \frac{x^{3}}{x^{3}+2 x^{2}-x-2} d x= \\
& =\int\left(1+\frac{-2 x^{2}+x+2}{x^{3}+2 x^{2}-x-2}\right) d x=\int d x+\int \frac{-2 x^{2}+x+2}{(x-1)(x+1)(x+2)} d x=
\end{aligned}
$$

Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

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

Add the first row to the third row:

$$
\left\{\begin{array}{l}
A+B+C=-2 \\
3 A+B=1 \\
3 A-B=0
\end{array}\right.
$$

Add the second row to the third row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B+C=-2 \\
3 A+B=1 \\
6 A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B+C=-2 \\
3 A+B=1 \\
A=\frac{1}{6}
\end{array}\right. \\
& \left\{\begin{array}{l}
B+C=-\frac{13}{6} \\
B=\frac{1}{2} \\
A=\frac{1}{6}
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-\frac{8}{3} \\
B=\frac{1}{2} \\
A=\frac{1}{6}
\end{array}\right. \\
& \frac{-2 x^{2}+x+2}{(x-1)(x+1)(x+2)}=\frac{1}{6(x-1)}+\frac{1}{2(x+1)}-\frac{8}{3(x+2)}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int d x+\int\left(\frac{1}{6(x-1)}+\frac{1}{2(x+1)}-\frac{8}{3(x+2)}\right) d x= \\
& =x+\frac{1}{6} \cdot \ln |x-1|+\frac{1}{2} \cdot \ln |x+1|-\frac{8}{3} \cdot \ln |x+2|+C
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_5-9"

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

- Last modified: 19:45, April 4, 2009.
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## Problem Kuznetsov Integrals 5-10

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