Show that

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

Let $x^{2}=t \Rightarrow 2 x d x=d t$

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

Let $\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}$

$1=A(t+3)+B(t+1)$   ...(1)

Substituting t = −3 and t = −1 in equation (1), we obtain

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

$=\frac{1}{2} \log |(t+1)|-\frac{1}{2} \log |t+3|+\mathrm{C}$

$=\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|+\mathrm{C}$

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