## Problem Statement

Find the indefinite integral:

$$
\int \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)} d x
$$
## Solution

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

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

$$
\begin{aligned}
& \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)^{2}}+\frac{B x+C}{x^{2}-x+1}= \\
& =\frac{A_{1}(x+1)\left(x^{2}-x+1\right)+A_{2}\left(x^{2}-x+1\right)+(B x+C)(x+1)^{2}}{(x+1)^{2}\left(x^{2}-x+1\right)}= \\
& =\frac{A_{1}\left(x^{3}+1\right)+A_{2}\left(x^{2}-x+1\right)+B\left(x^{3}+2 x^{2}+x\right)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}\left(x^{2}-x+1\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(A_{2}+2 B+C\right) x^{2}+\left(-A_{2}+B+2 C\right) x+\left(A_{1}+A_{2}+C\right)}{(x+1)^{2}\left(x^{2}-x+1\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=1 \\
A_{2}+2 B+C=2 \\
-A_{2}+B+2 C=10 \\
A_{1}+A_{2}+C=0
\end{array}\right.
\end{aligned}
$$

Subtract the second equation from the fourth:

$$
\left\{\begin{array}{l}
A_{1}+B=1 \\
A_{2}+2 B+C=2 \\
-A_{2}+B+2 C=10 \\
A_{1}-2 B=-2
\end{array}\right.
$$

Subtract the first equation from the fourth:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=1 \\
A_{2}+2 B+C=2 \\
-A_{2}+B+2 C=10 \\
-3 B=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}+C=0 \\
-A_{2}+2 C=9 \\
B=1
\end{array}\right.
\end{aligned}
$$

Add the second equation to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}+C=0 \\
3 C=9 \\
B=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
A_{2}=-3 \\
C=3 \\
B=1
\end{array}\right. \\
& \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)}=-\frac{3}{(x+1)^{2}}+\frac{x+3}{x^{2}-x+1}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)} d x=\int\left(-\frac{3}{(x+1)^{2}}+\frac{x+3}{x^{2}-x+1}\right) d x= \\
& =-\int \frac{3}{(x+1)^{2}} d x+\frac{1}{2} \cdot \int \frac{2 x-1+7}{x^{2}-x+1} d x= \\
& =\frac{3}{x+1}+\frac{1}{2} \cdot \int \frac{2 x-1}{x^{2}-x+1} d x+\frac{1}{2} \cdot \int \frac{7}{x^{2}-x+1} d x= \\
& =\frac{3}{x+1}+\frac{1}{2} \cdot \int \frac{d\left(x^{2}-x+1\right)}{x^{2}-x+1}+\frac{7}{2} \cdot \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}} d x= \\
& =\frac{3}{x+1}+\frac{1}{2} \cdot \ln \left(x^{2}-x+1\right)+\frac{7}{2} \cdot \int \frac{d\left(x-\frac{1}{2}\right)}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}= \\
& =\frac{3}{x+1}+\frac{1}{2} \cdot \ln \left(x^{2}-x+1\right)+\frac{7}{2} \cdot \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \cdot \operatorname{arctg}\left(\frac{x-\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)}\right)+C= \\
& =\frac{3}{x+1}+\frac{1}{2} \cdot \ln \left(x^{2}-x+1\right)+\frac{7}{\sqrt{3}} \cdot \operatorname{arctg}\left(\frac{2 x-1}{\sqrt{3}}\right)+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-13$ »

Categories: Kuznetsov's Integral Problems 7 | Integrals

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## Problem Kuznetsov Integrals 7-14

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