Question

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

Answer 1

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

$Let \ x^2 = t ⇒ 2x \ dx = dt$

∴ ∫$$\frac {2x}{(x^2+1)(x^2+3)}dx = $∫\frac {dt}{(t+1)(t+3)}$ ...(1)

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

$1 = A(t+3)+B(t+1)$ ...(1)

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

$A = \frac 12\ and \ B = -\frac 12$

∴ $\frac {1}{(t+1)(t+3)}$ = $\frac {1}{2(t+1)}-\frac {1}{2(t+3)}$

⇒ ∫$$\frac {2x}{(x^2+1)(x^2+3)}dx = ∫$$[\frac {1}{2(t+1)}-\frac {1}{2(t+3)}]dt

                                       =$\frac 12log\ |(t+1)|-\frac 12log\ |t+3|+C$

                                       =$\frac 12\ log|\frac {t+1}{t+3}|+C$

                                       =$\frac 12\ log|\frac {x^2+1}{x^2+3}|+C$