Let $I=\int\frac{2x}{(x^2+1)(x^2+3)}dx$ Put $x^2 = y \Rightarrow 2x\,dx = dy$ $\therefore I=\int\frac{dy}{(y+1)(y+3)}$ Let $\frac{1}{(y+1)(y+3)}=\frac{A}{y+1}+\frac{B}{y+3}$ $\Rightarrow 1 = A(y+3) + B(y+1)$ ...(i) Putting y = –1 in (i), we get $1=2A \Rightarrow A=\frac{1}{2}$Putting y = –3 in (i), we get $1=-2B \Rightarrow B=-\frac{1}{2}$$\Rightarrow \frac{1}{(y+1)(y+3)}=\frac{1}{2(y+1)}-\frac{1}{2(y+3)}$ $\therefore I=\int\left(\frac{1}{2(y+1)}-\frac{1}{2(y+3)}\right)dy$ $=\frac{1}{2}\int\frac{dy}{y+1}-\frac{1}{2}\int\frac{dy}{y+3}$$=\frac{1}{2}\log|y+1|-\frac{1}{2}\log|y+3|+C$$=\frac{1}{2}\log\left|\frac{y+1}{y+3}\right|+C$$=\frac{1}{2}\log\left|\frac{x^2+1}{x^2+3}\right|+C$