## Problem Statement

Find the indefinite integral:

$$
\int \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)} d x
$$
## Solution

$$
\int \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)} d x=
$$

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

$$
\begin{aligned}
& \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+x+1}= \\
& =\frac{(A x+B)\left(x^{2}+x+1\right)+(C x+D)\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}= \\
& =\frac{A x^{3}+A x^{2}+A x+B x^{2}+B x+B+C x^{3}+C x+D x^{2}+D}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}= \\
& =\frac{(A+C) x^{3}+(A+B+D) x^{2}+(A+B+C) x+(B+D)}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)} \\
& \left\{\begin{array}{l}
A+C=0 \\
A+B+D=4 \\
A+B+C=3 \\
B+D=4
\end{array}\right.
\end{aligned}
$$

Subtract the first equation from the third:

$$
\left\{\begin{array}{l}
A+C=0 \\
A+B+D=4 \\
B=3 \\
B+D=4
\end{array}\right.
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+C=0 \\
A+D=1 \\
B=3 \\
D=1
\end{array}\right. \\
& \left\{\begin{array}{l}
A+C=0 \\
A=0 \\
B=3 \\
D=1
\end{array}\right. \\
& \left\{\begin{array}{l}
C=0 \\
A=0 \\
B=3 \\
D=1
\end{array}\right. \\
& \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}=\frac{3}{x^{2}+1}+\frac{1}{x^{2}+x+1}
\end{aligned}
$$

Then:

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \ldots \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+7-21$ »

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

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## Problem Kuznetsov Integrals 7-22

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