Put $x^2=y$, so that $2 x d x=d y$
$\therefore \quad \int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x=\int \frac{d y}{(y+1)(y+3)}$
Let $\frac{1}{(y+1)(y+3)} \equiv \frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+3}$, we have
$1=\mathrm{A}(y+3)+\mathrm{B}(y+1) \quad \ldots(i)$
Put $\mathrm{y}=-1,-3$ in (i), we get; $\mathrm{A}=\frac{1}{2} \& \mathrm{~B}=-\frac{1}{2}$
$\begin{aligned}
&\therefore \int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x=\frac{1}{2} \int \frac{d y}{y+1}-\frac{1}{2} \int \frac{d y}{y+3} \\
&=\frac{1}{2} \log \left|\frac{x^2+1}{x^2+3}\right|+\mathrm{C}
\end{aligned}$