Since, $\int \frac{d x}{(x+2)\left(x^2+1\right)}$ $=a \log \left|1+x^2\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+C$
Suppose, $I=\int \frac{d x}{(x+2)\left(x^2+1\right)}$
Let, $\frac{1}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}$
$$
\therefore \quad 1=\mathrm{A}\left(\mathrm{x}^2+1\right)+(\mathrm{Bx}+\mathrm{C})(\mathrm{x}+2)
$$
$$
\Rightarrow \quad 1=(A+B) x^2+(2 B+C) x+A+2 C
$$
So, $\mathrm{A}+\mathrm{B}=0, \mathrm{~A}+2 \mathrm{C}=1,2 \mathrm{~B}+\mathrm{C}=0$
$\Rightarrow A=\frac{1}{5}, B=-\frac{1}{5}$ and $C=\frac{2}{5}$
Therefore, $\quad \int \frac{d x}{(x+2)\left(x^2+1\right)}$
$=\frac{1}{5} \int \frac{1}{x+2} d x+\int \frac{-\frac{1}{5} x+\frac{2}{5}}{x^2+1} d x$
$=\frac{1}{5} \int \frac{1}{x+2} d x-\frac{1}{5} \int \frac{x}{1+x^2} d x+\frac{1}{5} \int \frac{2}{1+x^2} d x$
$=\frac{1}{5} \log |x+2|-\frac{1}{10} \log \left|1+x^2\right|+\frac{2}{5} \tan ^{-1} x+C$
Hence, $a=\frac{-1}{10}, b=\frac{2}{5}$