Find: $\int \frac{2x}{(x^2 + 1)(x^2 - 4)} dx$

$\frac{1}{5} \ln \left| \frac{x^2 - 4}{x^2 + 1} \right| + C$

The correct answer is Option (2) → $\frac{1}{5} \ln \left| \frac{x^2 - 4}{x^2 + 1} \right| + C$

Let $I = \int \frac{2x dx}{(x^2 + 1)(x^2 - 4)}$

Put $x^2 = y ⇒2x dx = dy$

$I = \int \frac{dy}{(y + 1)(y - 4)}$

Let $\frac{1}{(y + 1)(y - 4)} = \frac{A}{(y + 1)} + \frac{B}{(y - 4)}$

$⇒A(y - 4) + B(y + 1) = 1$

$⇒(A + B)y - 4A + B = 1 + 0y$

After equating, we get:

$A + B = 0$ & $B - 4A = 1$

Solving we get, $A = -\frac{1}{5}$ & $B = \frac{1}{5}$

Put value of A & B in I, we get:

$I = \int \frac{-1}{5(y + 1)} dy + \int \frac{1}{5(y - 4)} dy$

$= -\frac{1}{5} \log |y + 1| + \frac{1}{5} \log |y - 4| + C$

$= -\frac{1}{5} \log |x^2 + 1| + \frac{1}{5} \log |x^2 - 4| + C$

$= \frac{1}{5} \log \left| \frac{x^2 - 4}{x^2 + 1} \right| + C$