## Problem Statement

Find the indefinite integral:

$$
\int \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} d x
$$
## Solution

$$
\int \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)}=\frac{A x+B}{x^{2}+x+3}+\frac{C x+D}{x^{2}+2 x+3}= \\
& =\frac{(A x+B)\left(x^{2}+2 x+3\right)+(C x+D)\left(x^{2}+x+3\right)}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)}= \\
& =\frac{A x^{3}+2 A x^{2}+3 A x+B x^{2}+2 B x+3 B+C x^{3}+C x^{2}+3 C x+D x^{2}+D x+3 D}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)}= \\
& =\frac{(A+C) x^{3}+(2 A+B+C+D) x^{2}+(3 A+2 B+3 C+D) x+(3 B+3 D)}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} \\
& \left\{\begin{array}{l}
A+C=3 \\
2 A+B+C+D=7 \\
3 A+2 B+3 C+D=12 \\
3 B+3 D=6
\end{array}\right. \\
& \left\{\begin{array}{l}
A+C=3 \\
2 A+B+C+D=7 \\
3 A+2 B+3 C+D=12 \\
B+D=2
\end{array}\right.
\end{aligned}
$$

Subtract the fourth equation from the second:

$$
\left\{\begin{array}{l}
A+C=3 \\
2 A+C=5 \\
3 A+2 B+3 C+D=12 \\
B+D=2
\end{array}\right.
$$

Subtract the first equation from the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+C=3 \\
A=2 \\
3 A+2 B+3 C+D=12 \\
B+D=2
\end{array}\right. \\
& \left\{\begin{array}{l}
C=1 \\
A=2 \\
2 B+3 C+D=6 \\
B+D=2
\end{array}\right. \\
& \left\{\begin{array}{l}
C=1 \\
A=2 \\
2 B+D=3 \\
B+D=2
\end{array}\right.
\end{aligned}
$$

Subtract the fourth equation from the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
C=1 \\
A=2 \\
B=1 \\
B+D=2
\end{array}\right. \\
& \left\{\begin{array}{l}
C=1 \\
A=2 \\
B=1 \\
D=1
\end{array}\right. \\
& \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)}=\frac{2 x+1}{x^{2}+x+3}+\frac{x+1}{x^{2}+2 x+3}
\end{aligned}
$$

Then:

$$
\begin{aligned}
& \int \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} d x=\int\left(\frac{2 x+1}{x^{2}+x+3}+\frac{x+1}{x^{2}+2 x+3}\right) d x= \\
& =\int \frac{d\left(x^{2}+x+3\right)}{x^{2}+x+3} d x+\frac{1}{2} \cdot \int \frac{2 x+2}{x^{2}+2 x+3} d x= \\
& =\ln \left(x^{2}+x+3\right)+\frac{1}{2} \cdot \int \frac{d\left(x^{2}+2 x+3\right)}{x^{2}+2 x+3} d x= \\
& =\ln \left(x^{2}+x+3\right)+\frac{1}{2} \cdot \ln \left(x^{2}+2 x+3\right)+C
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_7-30$ »

Categories: Kuznetsov's Integral Problems 7 | Integrals

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- Last edited: 09:39, 22 April 2009.
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