Question

The value of the integral $\int \frac{2x}{(x^2+1)(x^2+3)}dx$ is
(where $C$ is an arbitrary constant)

• A. $\frac{1}{2}\ln \left|\frac{x^2+1}{x^2-3}\right|+C$
• B. $\frac{1}{2}\ln \left|\frac{x^2-1}{x^2-3}\right|+C$
• C. $\frac{1}{2}\ln \left|\frac{x^2+1}{x^2+3}\right|+C$
• D. $\frac{1}{2}\ln \left|\frac{x^2-1}{x^2+3}\right|+C$

Answer 1

The correct option is C $\frac{1}{2}\ln \left|\frac{x^2+1}{x^2+3}\right|+C$

$\frac{2x}{(x^2+1)(x^2+3)}$
Let $x^2=t\Rightarrow 2xdx=dt$
$\therefore I=\int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \frac{dt}{(t+1)(t+3)}\cdots \left(1\right)$
Let $\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}$
$1=A(t+3)+B(t+1)\cdots \left(2\right)$
Equating the coefficients of $t$ and constant, we obtain
$A+B=0$ and $3A+B=1$
On solving, we obtain
$A=\frac{1}{2}$ and $B=-\frac{1}{2}$
$\therefore \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\Rightarrow I=\int [\frac{1}{2(t+1)}-\frac{1}{2(t+3)}]dt=\frac{1}{2}\ln \left|\right(t+1\left)\right|-\frac{1}{2}\ln |t+3|+C=\frac{1}{2}\ln \left|\frac{t+1}{t+3}\right|+C=\frac{1}{2}\ln \left|\frac{x^2+1}{x^2+3}\right|+C$