Find $\int \frac{2x}{(x^2 + 1)(x^2 + 2)} dx$

$\ln \left| \frac{x^2 + 1}{x^2 + 2} \right| + C$

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 $B = -1$

$∴I = \int \left( \frac{1}{x^2 + 1} - \frac{1}{x^2 + 2} \right) 2x dx$

$I = \int \frac{2x}{x^2 + 1} dx - \int \frac{2x}{x^2 + 2} dx$

$I = \ln |x^2 + 1| - \ln |x^2 + 2| + C$

$I= \ln \left| \frac{x^2 + 1}{x^2 + 2} \right| + C$