Question

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

Answer 1

$\int \frac{2x}{(x^2+1)(x^2+3)}.dx$
Let $x^2=t$
Differentiate both sides w.r.t. $x$
$\Rightarrow 2xdx=dt$
$\int \frac{2x}{(x^2+1)(x^2+3)}.dx$
$=\int \frac{dt}{(t+1)(t+3)}$
Using partial fraction
$\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}$
$\Rightarrow 1=A(t+3)+B(t+1)$
Putting $t=-1$
$\Rightarrow 1=A(-1+3)+B(-1+1)$
$\Rightarrow A=\frac{1}{2}$
Similarly putting $t=-3$
$\Rightarrow 1=A(-3+3)+B(-3+1)$
$\Rightarrow B=-\frac{1}{2}$
Now,
$\int \frac{1}{(t+1)(t+3)}.dt$
$=\int \frac{1}{2(t+1)}.dt+\int \frac{-1}{2(t+3)}.dt$
$=\frac{1}{2}\int \frac{1}{(t+1)}.dt-\frac{1}{2}\int \frac{1}{(t+3)}.dt$
$=\frac{1}{2}\log \left|\begin{array}{c}t+1\end{array}\right|-\frac{1}{2}\log \left|\begin{array}{c}t+3\end{array}\right|+C$
$=\frac{1}{2}\log \left|\begin{array}{c}\frac{t+1}{t+3}\end{array}\right|+C$
$=\frac{1}{2}\log \left|\begin{array}{c}\frac{x^2+1}{x^2+3}\end{array}\right|+C$
$[\because t=x^2]$
Where $C$ is constant of integration.