## Problem Statement

Calculate the indefinite integral:

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

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## Solution

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

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

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

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

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=-6 \\
12 A-4 B_{1}+B_{3}=13 \\
32 A-16 B_{1}-4 B_{2}=20
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=-6 \\
12 A-4 B_{1}+B_{3}=13 \\
56 A-8 B_{1}=-4
\end{array}\right.
$$

Add the first equation multiplied by 8 to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
6 A+2 B_{1}+B_{2}=-6 \\
12 A-4 B_{1}+B_{3}=13 \\
64 A=4
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=\frac{15}{16} \\
2 B_{1}+B_{2}=-\frac{51}{8} \\
-4 B_{1}+B_{3}=\frac{49}{4} \\
A=\frac{1}{16}
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=\frac{15}{16} \\
B_{2}=-\frac{33}{4} \\
B_{3}=16 \\
A=\frac{1}{16}
\end{array}\right. \\
& \frac{x^{3}-6 x^{2}+13 x-6}{(x-2)(x+2)^{3}}=\frac{1}{16(x-2)}+\frac{15}{16(x+2)}-\frac{33}{4(x+2)^{2}}+\frac{16}{(x+2)^{3}}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{x^{3}-6 x^{2}+13 x-6}{(x-2)(x+2)^{3}} d x=\int\left(\frac{1}{16(x-2)}+\frac{15}{16(x+2)}-\frac{33}{4(x+2)^{2}}+\frac{16}{(x+2)^{3}}\right) d x= \\
& =\frac{1}{16} \cdot \ln |x-2|+\frac{15}{16} \cdot \ln |x+2|+\frac{33}{4(x+2)}-\frac{8}{(x+2)^{2}}+C= \\
& =\frac{1}{16} \cdot \ln |x-2|+\frac{15}{16} \cdot \ln |x+2|+\frac{33 x+34}{4(x+2)^{2}}+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}$ 6-30»

Categories: Problems with errors in solution | Kuznetsov's Integral Problems 6 | Integrals

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## Problem Kuznetsov Integrals 6-31

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