Find $\int \frac{3x-2}{(x+1)^2(x+3)} \, dx$

$\frac{11}{4} \ln\left| \frac{x + 1}{x + 3} \right| + \frac{5}{2(x + 1)} + C$

The correct answer is Option (4) → $\frac{11}{4} \ln\left| \frac{x + 1}{x + 3} \right| + \frac{5}{2(x + 1)} + C$

We write

$\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$

so that

$3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$

$= A(x^2 + 4x + 3) + B(x+3) + C(x^2 + 2x + 1)$

Comparing coefficient of $x^2$, $x$ and constant term on both sides, we get $A + C = 0$, $4A + B + 2C = 3$ and $3A + 3B + C = -2$. Solving these equations, we get $A = \frac{11}{4}$, $B = \frac{-5}{2}$ and $C = \frac{-11}{4}$. Thus the integrand is given by

$\frac{3x-2}{(x+1)^2(x+3)} = \frac{11}{4(x+1)} - \frac{5}{2(x+1)^2} - \frac{11}{4(x+3)}$

$\int \frac{3x-2}{(x+1)^2(x+3)} = \frac{11}{4} \int \frac{dx}{x+1} - \frac{5}{2} \int \frac{dx}{(x+1)^2} - \frac{11}{4} \int \frac{dx}{x+3}$

$= \frac{11}{4} \log |x+1| + \frac{5}{2(x+1)} - \frac{11}{4} \log |x+3| + C$

$= \frac{11}{4} \log \left| \frac{x+1}{x+3} \right| + \frac{5}{2(x+1)} + C$