$\cfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Let $I = \int {\cfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}dx}$

Put ${x^2} = y$ $\Rightarrow$ $2xdx = dy$

$\therefore$ $I = \int {\cfrac{{dy}}{{\left( {y + 1} \right)\left( {y + 3} \right)}}}$

We write, $\cfrac{1}{{\left( {y + 1} \right)\left( {y + 3} \right)}} = \cfrac{A}{{y + 1}} + \cfrac{B}{{y + 3}}$

$1 = A\left( {y + 3} \right) + B\left( {y + 1} \right)$

…(i)
Putting $y = - 1$ in $(i)$ ,

we get $1 = 2A$ $\Rightarrow$ $A = \cfrac{1}{2}$

Putting $y = - 3$ in $(i)$ ,

we get $1 = - 2B$ $\Rightarrow$ $B = - \cfrac{1}{2}$

$\Rightarrow$ $\cfrac{1}{{\left( {y + 1} \right)\left( {y + 3} \right)}} = \cfrac{1}{{2\left( {y + 1} \right)}} - \cfrac{1}{{2\left( {y + 3} \right)}}$

$\therefore$ $I = \int {\left[ {\cfrac{1}{{2\left( {y + 1} \right)}} - \cfrac{1}{{2\left( {y + 3} \right)}}} \right]} dy = \cfrac{1}{2}\int {\cfrac{{dy}}{{y + 1}}} - \cfrac{1}{2}\int {\cfrac{{dy}}{{y + 3}}}$

$= \cfrac{1}{2}\log \left| {y + 1} \right| - \cfrac{1}{2}\log \left| {y + 3} \right| + C$

$= \cfrac{1}{2}\log \left| {\cfrac{{y + 1}}{{y + 3}}} \right| + C = \cfrac{1}{2}\log \left| {\cfrac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C$