## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}} d x
$$
## Solution

$$
\int \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}} d x=
$$

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

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

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

$$
\left\{\begin{array}{l}
A+B_{1}=2 \\
-3 A+B_{2}=-6 \\
3 A-3 B_{1}+B_{2}+B_{3}=7 \\
-7 A-4 B_{1}-4 B_{2}=-14
\end{array}\right.
$$

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

$$
\left\{\begin{array}{l}
A+B_{1}=2 \\
-3 A+B_{2}=-6 \\
3 A-3 B_{1}+B_{2}+B_{3}=7 \\
-19 A-4 B_{1}=-38
\end{array}\right.
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B_{1}=2 \\
-3 A+B_{2}=-6 \\
3 A-3 B_{1}+B_{2}+B_{3}=7 \\
-15 A=-30
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
-3 B_{1}+B_{2}+B_{3}=1 \\
A=2
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=0 \\
B_{2}=0 \\
B_{3}=1 \\
A=2
\end{array}\right. \\
& \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}}=\frac{2}{x+2}+\frac{1}{(x-1)^{3}}
\end{aligned}
$$

Then:

$$
\int \frac{2 x^{3}-6 x^{2}+7 x}{(x+2)(x-1)^{3}} d x=\int\left(\frac{2}{x+2}+\frac{1}{(x-1)^{3}}\right) d x=2 \cdot \ln |x+2|-\frac{1}{2(x-1)^{2}}+C
$$

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 $\% \mathrm{D} 1 \% 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}+6-28 »

Categories: Kuznetsov's Problem Book Integrals Problem $6 \mid$ Integrals

- Last edited: 08:43, April 10, 2009.
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## Problem Kuznetsov Integrals 6-29

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