## Problem Statement

Find the indefinite integral:

$$
\int \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)} d x
$$
## Solution

$$
\int \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)}=\frac{A_{1}}{x-2}+\frac{A_{2}}{(x-2)^{2}}+\frac{B x+C}{x^{2}-x+1}= \\
& =\frac{A_{1}(x-2)\left(x^{2}-x+1\right)+A_{2}\left(x^{2}-x+1\right)+(B x+C)(x-2)^{2}}{(x-2)^{2}\left(x^{2}-x+1\right)}= \\
& =\frac{A_{1}\left(x^{3}-3 x^{2}+3 x-2\right)+A_{2}\left(x^{2}-x+1\right)+B\left(x^{3}-4 x^{2}+4 x\right)+C\left(x^{2}-4 x+4\right)}{(x-2)^{2}\left(x^{2}-x+1\right)}= \\
& =\frac{\left(A_{1}+B\right) x^{3}+\left(-3 A_{1}+A_{2}-4 B+C\right) x^{2}+\left(3 A_{1}-A_{2}+4 B-4 C\right) x+\left(-2 A_{1}+A_{2}+4 C\right)}{(x-2)^{2}\left(x^{2}-x+1\right)} \\
& \left\{\begin{array}{l}
A_{1}+B=-3 \\
-3 A_{1}+A_{2}-4 B+C=13 \\
3 A_{1}-A_{2}+4 B-4 C=-13 \\
-2 A_{1}+A_{2}+4 C=1
\end{array}\right.
\end{aligned}
$$

Add the second equation to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=-3 \\
-3 A_{1}+A_{2}-4 B+C=13 \\
-3 C=0 \\
-2 A_{1}+A_{2}+4 C=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}+B=-3 \\
-3 A_{1}+A_{2}-4 B=13 \\
C=0 \\
-2 A_{1}+A_{2}=1
\end{array}\right.
\end{aligned}
$$

Subtract the fourth equation from the second:

$$
\left\{\begin{array}{l}
A_{1}+B=-3 \\
-A_{1}-4 B=12 \\
C=0 \\
-2 A_{1}+A_{2}=1
\end{array}\right.
$$

Add the first equation to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A_{1}+B=-3 \\
-3 B=9 \\
C=0 \\
-2 A_{1}+A_{2}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
B=-3 \\
C=0 \\
-2 A_{1}+A_{2}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A_{1}=0 \\
B=-3 \\
C=0 \\
A_{2}=1
\end{array}\right. \\
& \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)}=\frac{1}{(x-2)^{2}}-\frac{3 x}{x^{2}-x+1}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)} d x=\int\left(\frac{1}{(x-2)^{2}}-\frac{3 x}{x^{2}-x+1}\right) d x= \\
& =\int \frac{1}{(x-2)^{2}} d x-\frac{3}{2} \cdot \int \frac{2 x-1+1}{x^{2}-x+1} d x= \\
& =-\frac{1}{x-2}-\frac{3}{2} \cdot \int \frac{2 x-1}{x^{2}-x+1} d x-\frac{3}{2} \cdot \int \frac{1}{x^{2}-x+1} d x= \\
& =-\frac{1}{x-2}-\frac{3}{2} \cdot \int \frac{d\left(x^{2}-x+1\right)}{x^{2}-x+1}-\frac{3}{2} \cdot \int \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}} d x= \\
& =-\frac{1}{x-2}-\frac{3}{2} \cdot \ln \left(x^{2}-x+1\right)-\frac{3}{2} \cdot \int \frac{d\left(x-\frac{1}{2}\right)}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}= \\
& =-\frac{1}{x-2}-\frac{3}{2} \cdot \ln \left(x^{2}-x+1\right)-\frac{3}{2} \cdot \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \cdot \operatorname{arctg}\left(\frac{x-\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)}\right)+C= \\
& =-\frac{1}{x-2}-\frac{3}{2} \cdot \ln \left(x^{2}-x+1\right)-\sqrt{3} \cdot \operatorname{arctg}\left(\frac{2 x-1}{\sqrt{3}}\right)+C
\end{aligned}
$$

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\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_7-12$ »

Categories: Kuznetsov's Problem Book Integrals Problem 7 | Integrals

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## Problem Kuznetsov Integrals $7-13$

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