Evaluate the following integral:
$\int \frac{3 x-2}{(x+1)^{2}(x+3)}$
$I=\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x$
$\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{c}{x+3}$
$3 x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^{2}$
Put $x=-1$
$-3-2=A \times 0+B \times(-1+3)+C \times 0$
$-5=2 B$
$B=-\frac{5}{2}$
Put $x=-3$
$-9-2=C \times(-2)(-2)$
$-11=4 C$
$C=-\frac{11}{4}$
Equating coefficients of constants
$-2=3 A+3 B+C$
$-2=3 A+3 \times \frac{-5}{2}-\frac{11}{4}$
$A=\frac{11}{4}$
Thus,
$I=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
$I=\frac{11}{4} \log |x+1|-\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+C$
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