Evaluate the following integral:

Solution:

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

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

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

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,

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

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