Question

Integrate the rational function $\frac{2x}{(x^2+1)(x^2+3)}$

Answer 1

Let $x^2=t\Rightarrow 2xdx=dt$
$\therefore \int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \frac{dt}{(t+1)(t+3)}$ .........(1)
Let $\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}$
$\Rightarrow 1=A(t+3)+b(t+1)$ ......... (1)
Substituting $t=-3$ and $t=-1$ in equation (1) we obtain
$A=\frac{1}{2}$ and $B=-\frac{1}{2}$
$\therefore \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}$
$\Rightarrow \int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \{\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\}dt$
$=\frac{1}{2}\log |(t+1)|-\frac{1}{2}\log |t+3|+C$
$=\frac{1}{2}\log \left|\frac{t+1}{t+3}\right|+C$
$=\frac{1}{2}\log \left|\frac{x^2+1}{x^2+3}\right|+C$