## Problem Statement

Find the indefinite integral:

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

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

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

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

Subtract the first equation from the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+C=2 \\
B+C+D=3 \\
D=1 \\
B+D=2
\end{array}\right. \\
& \left\{\begin{array}{l}
A+C=2 \\
B+C=2 \\
D=1 \\
B=1
\end{array}\right.
\end{aligned}
$$

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

Then:

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

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-16$ "

Categories: Kuznetsov's Integral Problems 7 | Integrals

- Last edited: 09:04, April 21, 2009.
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## Problem Kuznetsov Integrals $7-17$

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