## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}} d x
$$
## Solution

$$
\int \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}} d x=
$$

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

$$
\begin{aligned}
& \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}}=\frac{A}{x+2}+\frac{B_{1}}{x-2}+\frac{B_{2}}{(x-2)^{2}}+\frac{B_{3}}{(x-2)^{3}}= \\
& =\frac{A(x-2)^{3}+B_{1}(x+2)(x-2)^{2}+B_{2}(x+2)(x-2)+B_{3}(x+2)}{(x+2)(x-2)^{3}}= \\
& =\frac{A\left(x^{3}-6 x^{2}+12 x-8\right)+B_{1}\left(x^{3}-2 x^{2}-4 x+8\right)+B_{2}\left(x^{2}-4\right)+B_{3}(x+2)}{(x+2)(x-2)^{3}}= \\
& =\frac{\left(A+B_{1}\right) x^{3}+\left(-6 A-2 B_{1}+B_{2}\right) x^{2}+\left(12 A-4 B_{1}+B_{3}\right) x+\left(-8 A+8 B_{1}-4 B_{2}\right.}{(x+2)(x-2)^{3}} \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
-6 A-2 B_{1}+B_{2}=-6 \\
12 A-4 B_{1}+B_{3}=11 \\
-8 A+8 B_{1}-4 B_{2}+2 B_{3}=-10
\end{array}\right.
\end{aligned}
$$

Add the first equation multiplied by 6 to the second equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
4 B_{1}+B_{2}=0 \\
12 A-4 B_{1}+B_{3}=11 \\
-8 A+8 B_{1}-4 B_{2}+2 B_{3}=-10
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
B_{2}=-4 B_{1} \\
12 A-4 B_{1}+B_{3}=11 \\
-8 A+24 B_{1}+2 B_{3}=-10
\end{array}\right.
\end{aligned}
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
B_{2}=-4 B_{1} \\
12 A-4 B_{1}+B_{3}=11 \\
-32 A+32 B_{1}=-32
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B_{1}=1 \\
B_{2}=-4 B_{1} \\
12 A-4 B_{1}+B_{3}=11 \\
A-B_{1}=1
\end{array}\right.
\end{aligned}
$$

Add the first equation to the fourth equation:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=1 \\
B_{2}=-4 B_{1} \\
12 A-4 B_{1}+B_{3}=11 \\
2 A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=-1 \\
A=1
\end{array}\right. \\
& \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}}=\frac{1}{x+2}-\frac{1}{(x-2)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{x^{3}-6 x^{2}+11 x-10}{(x+2)(x-2)^{3}} d x=\int\left(\frac{1}{x+2}-\frac{1}{(x-2)^{3}}\right) d x=\ln |x+2|+\frac{1}{2(x-2)^{2}}+C
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_6-5$ »

Categories: Kuznetsov's Integral Problems 6-6

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