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

$\frac{3}{4} \ln \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2(x + 1)} + C$

The correct answer is Option (2) → $\frac{3}{4} \ln \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2(x + 1)} + C$

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

Let $\frac{2x + 1}{(x + 1)^2(x - 1)} = \frac{A}{(x + 1)} + \frac{B}{(x + 1)^2} + \frac{C}{(x - 1)}$

$2x + 1 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)^2$

$2x + 1 = A(x^2 - 1) + Bx - B + C(x^2 + 2x + 1)$

On comparing, we get

$A + C = 0$; $B + 2C = 2$; and $-A - B + C = 1$

On solving, we get

$A = -\frac{3}{4}$, $B = \frac{1}{2}$, and $C = \frac{3}{4}$

$∴\frac{2x + 1}{(x + 1)^2(x - 1)} = \frac{-3}{4(x + 1)} + \frac{1}{2(x + 1)^2} + \frac{3}{4(x - 1)}$

$I = \frac{3}{4}\int \frac{1}{x - 1} dx - \frac{3}{4}\int \frac{1}{x + 1} dx + \frac{1}{2}\int \frac{1}{(x + 1)^2} dx$

$I = \frac{3}{4} \ln |x - 1| - \frac{3}{4} \ln |x + 1| + \frac{1}{2} \left( -\frac{1}{x + 1} \right) + C$

$I = \frac{3}{4} \ln |x - 1| - \frac{3}{4} \ln |x + 1| - \frac{1}{2(x + 1)} + C$

$I = \frac{3}{4} \ln \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2(x + 1)} + C$