Find $\int \frac{2x}{(x^2 + 1)(x^2 + 2)} dx$
Answer & explanation
Correct answer: option 3
Let $I = \int \frac{2x}{(x^2 + 1)(x^2 + 2)} dx$
Let $\frac{1}{(x^2 + 1)(x^2 + 2)} = \frac{A}{(x^2 + 1)} + \frac{B}{(x^2 + 2)}$
$⇒1 = A(x^2 + 2) + B(x^2 + 1)$
$⇒1 = (A + B)x^2 + (2A + B)$
On comparing both sides, we get
$A + B = 0$ and $2A + B =0$
On solving the above equations, we get
$∴ A = 1$ and $B = -1$
$∴I = \int \left( \frac{1}{x^2 + 1} - \frac{1}{x^2 + 2} \right) 2x dx$
$I = \int \frac{2x}{x^2 + 1} dx - \int \frac{2x}{x^2 + 2} dx$
$I = \ln |x^2 + 1| - \ln |x^2 + 2| + C$
$I= \ln \left| \frac{x^2 + 1}{x^2 + 2} \right| + C$