Evaluate the following integral:

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

$\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+4}$

$1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$

$=(A+C) x^{3}+(B+D) x^{2}+(4 A+C) x+4 B+D$

Equating similar terms

$A+C=0$

$B+D=0$

$4 A+C=0$

$4 B+D=1$

We get, $\mathrm{A}=0 \mathrm{~B}=\frac{1}{3} \mathrm{C}=0 \mathrm{D}=-\frac{1}{3}$

Thus,

$I=\int \frac{\frac{1}{3} d x}{x^{2}+1}-\int \frac{\frac{1}{3} d x}{x^{2}+4}$

$=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$