## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}}=\frac{A}{x-2}+\frac{B_{1}}{x+2}+\frac{B_{2}}{(x+2)^{2}}+\frac{B_{3}}{(x+2)^{3}}= \\
& =\frac{A(x+2)^{3}+B_{1}(x-2)(x+2)^{2}+B_{2}(x-2)(x+2)+B_{3}(x-2)}{(x-2)(x+2)^{3}}= \\
& =\frac{A\left(x^{3}+6 x^{2}+12 x+8\right)+B_{1}\left(x^{3}+2 x^{2}-4 x-8\right)+B_{2}\left(x^{2}-4\right)+B_{3}(x-2)}{(x-2)(x+2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(6 A+2 B_{1}+B_{2}\right) x^{2}+\left(12 A-4 B_{1}+B_{3}\right) x+\left(8 A-8 B_{1}-4 B_{2}-2 B_{3}\right)}{(x-2)(x+2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=6 \\
12 A-4 B_{1}+B_{3}=18 \\
8 A-8 B_{1}-4 B_{2}-2 B_{3}=-4
\end{array}\right.
\end{aligned}
$$

Add the third equation multiplied by 2 to the fourth equation:

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=6 \\
12 A-4 B_{1}+B_{3}=18 \\
32 A-16 B_{1}-4 B_{2}=32
\end{array}\right.
$$

Add the second equation multiplied by 4 to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=6 \\
12 A-4 B_{1}+B_{3}=18 \\
56 A-8 B_{1}=56
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=6 \\
12 A-4 B_{1}+B_{3}=18 \\
7 A-B_{1}=7
\end{array}\right.
\end{aligned}
$$

Add the first equation to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=6 \\
12 A-4 B_{1}+B_{3}=18 \\
8 A=8
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
2 B_{1}+B_{2}=0 \\
-4 B_{1}+B_{3}=6 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=6 \\
A=1
\end{array}\right. \\
& \frac{x^{3}+6 x^{2}+4 x+24}{(x-2)(x+2)^{3}}=\frac{1}{x-2}+\frac{6}{(x+2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}+6 x^{2}+18 x-4}{(x-2)(x+2)^{3}} d x=\int\left(\frac{1}{x-2}+\frac{6}{(x+2)^{3}}\right) d x=\ln |x-2|-\frac{3}{(x+2)^{2}}+C
$$

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_6-23» Categories: Kuznetsov's Integral Problems 6-23

- Last edited: 21:00, April 9, 2009.
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## Problem Kuznetsov Integrals 6-24

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