## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}}=\frac{A}{x+1}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}{(x-2)^{3}}= \\
& =\frac{A(x-2)^{3}+B_{1}(x+1)(x-2)^{2}+B_{2}(x+1)(x-2)+B_{3}(x+1)}{(x+1)(x-2)^{3}}= \\
& =\frac{A\left(x^{3}-6 x^{2}+12 x-8\right)+B_{1}\left(x^{3}-3 x^{2}+4\right)+B_{2}\left(x^{2}-x-2\right)+B_{3}(x+1)}{(x+1)(x-2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(-6 A-3 B_{1}+B_{2}\right) x^{2}+\left(12 A-B_{2}+B_{3}\right) x+\left(-8 A+4 B_{1}-2 B_{2}+B_{3}\right)}{(x+1)(x-2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-3 B_{1}+B_{2}=-6 \\
12 A-B_{2}+B_{3}=14 \\
-8 A+4 B_{1}-2 B_{2}+B_{3}=-6
\end{array}\right.
\end{aligned}
$$

Subtract the fourth equation from the third:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-3 B_{1}+B_{2}=-6 \\
20 A-4 B_{1}+B_{2}=20 \\
-8 A+4 B_{1}-2 B_{2}+B_{3}=-6
\end{array}\right.
$$

Subtract the second equation from the third:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-3 B_{1}+B_{2}=-6 \\
26 A-B_{1}=26 \\
-8 A+4 B_{1}-2 B_{2}+B_{3}=-6
\end{array}\right.
$$

Add the first equation to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-3 B_{1}+B_{2}=-6 \\
27 A=27 \\
-8 A+4 B_{1}-2 B_{2}+B_{3}=-6
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
-3 B_{1}+B_{2}=0 \\
A=1 \\
4 B_{1}-2 B_{2}+B_{3}=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
A=1 \\
-2 B_{2}+B_{3}=2 \\
B_{1}=0 \\
B_{2}=0 \\
A=1 \\
B_{3}=2
\end{array}\right. \\
& \frac{x^{3}-6 x^{2}+13 x-7}{(x+1)(x-2)^{3}}=\frac{1}{x+1}+\frac{2}{(x-2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}-6 x^{2}+14 x-6}{(x+1)(x-2)^{3}} d x=\int\left(\frac{1}{x+1}+\frac{2}{(x-2)^{3}}\right) d x=\ln |x+1|-\frac{1}{(x-2)^{2}}+C
$$

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_6-12» Categories: Kuznetsov's Problem Book Integrals Problem $6 \mid$ Integrals

- Last edited: 08:23, April 9, 2009.
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## Problem Kuznetsov Integrals 6-13

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