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

(N/A) The integrand is of the type as given in the table below:

Form of the rational function	Form of the partial fraction
$\frac{p x^{2}+q x+r}{(x-a)^{2}(x-b)}$	$\frac{A}{x-a}+\frac{B}{(x-a)^{2}}+\frac{C}{x-b}$

We write: $\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+3}$
Multiplying by the denominator,we get: $3 x-2 = A(x+1)(x+3)+B(x+3)+C(x+1)^{2}$
Expanding the terms: $3 x-2 = A(x^{2}+4 x+3)+B(x+3)+C(x^{2}+2 x+1)$
Comparing the coefficients of $x^{2}, x$ and the constant term on both sides,we get:
$A+C=0$
$4 A+B+2 C=3$
$3 A+3 B+C=-2$
Solving these equations,we get $A=\frac{11}{4}, B=\frac{-5}{2}$ and $C=\frac{-11}{4}$.
Thus,the integrand is: $\frac{3 x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}$
Integrating both sides:
$\int \frac{3 x-2}{(x+1)^{2}(x+3)} d x = \frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
$= \frac{11}{4} \log |x+1| + \frac{5}{2(x+1)} - \frac{11}{4} \log |x+3| + C$
$= \frac{11}{4} \log \left|\frac{x+1}{x+3}\right| + \frac{5}{2(x+1)} + C$