. $\cfrac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}$

: Let $I = \int {\cfrac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}dx}$

Now, consider $\cfrac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}$ Put${x^2} = t$

We write, $\cfrac{1}{{\left( {t + 1} \right)\left( {t + 4} \right)}} = \cfrac{A}{{\left( {t + 1} \right)}} + \cfrac{B}{{t + 4}}$

$\Rightarrow$ $1 = A\left( {t + 4} \right) + B\left( {t + 1} \right)$

….(i)
Putting $t = - 1$ in (i),

we get $1 = A\left( { - 1 + 4} \right)$ $\Rightarrow$ $A = \cfrac{1}{3}$

Putting $t = - 4$ in (i),

we get $1 = B\left( { - 4 + 1} \right)$ $\Rightarrow$ $B = - \cfrac{1}{3}$

$\therefore$ $\cfrac{1}{{\left( {t + 1} \right)\left( {t + 4} \right)}} = \cfrac{1}{{3\left( {t + 1} \right)}} - \cfrac{1}{{3\left( {t + 4} \right)}} = \cfrac{1}{{3\left( {{x^2} + 1} \right)}} - \cfrac{1}{{3\left( {{x^2} + 4} \right)}}$

Now, $I = \int {\left[ {\cfrac{1}{{3\left( {{x^2} + 1} \right)}} - \cfrac{1}{{3\left( {{x^2} + 4} \right)}}} \right]} dx$

$= \left( {\cfrac{1}{3}{{\tan }^{ - 1}}x} \right) - \left( {\cfrac{1}{3} \times \cfrac{1}{2}{{\tan }^{ - 1}}\left( {\cfrac{x}{2}} \right)} \right) + C$

$= \cfrac{1}{3}{\tan ^{ - 1}}x - \cfrac{1}{6}{\tan ^{ - 1}}\left( {\cfrac{x}{2}} \right) + C$