## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}}=\frac{A}{x}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}{(x-2)^{3}}= \\
& =\frac{A(x-2)^{3}+B_{1} x(x-2)^{2}+B_{2} x(x-2)+B_{3} x}{x(x-2)^{3}}= \\
& =\frac{A\left(x^{3}-6 x^{2}+12 x-8\right)+B_{1}\left(x^{3}-4 x^{2}+4 x\right)+B_{2}\left(x^{2}-2 x\right)+B_{3} x}{x(x-2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(-6 A-4 B_{1}+B_{2}\right) x^{2}+\left(12 A+4 B_{1}-2 B_{2}+B_{3}\right) x-8 A}{x(x-2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-4 B_{1}+B_{2}=-6 \\
12 A+4 B_{1}-2 B_{2}+B_{3}=13 \\
-8 A=-8
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
-4 B_{1}+B_{2}=0 \\
4 B_{1}-2 B_{2}+B_{3}=1 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
-2 B_{2}+B_{3}=1 \\
A=1
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1 \\
A=1
\end{array}\right. \\
& \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}}=\frac{1}{x}+\frac{1}{(x-2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}-6 x^{2}+13 x-8}{x(x-2)^{3}} d x=\int\left(\frac{1}{x}+\frac{1}{(x-2)^{3}}\right) d x=\ln |x|-\frac{1}{2(x-2)^{2}}+C
$$

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_6-11" Categories: Kuznetsov's Integral Problems 6-11 | Integrals

- Last edited: 08:03, April 9, 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 6-11

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