Evaluate the following integral:

Solution:

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

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

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

$2 x+1=A x^{2}-3 A x+9 A+B x^{2}-5 B x-6 B+C x+2 C$

Put $x=3$

$7=5 C$

$C=\frac{7}{5}$

Put $x=-2$

$-3=0 A$

$-11=4 C$

$C=-\frac{11}{4}$

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,

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

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