Evaluate the following integral:
$\int \frac{2 x+1}{(x+2)(x-3)^{2}}$
$I=\int \frac{2 x+1}{(x+2)(x-3)^{2}} d x$
$\frac{2 x+1}{(x+2)(x-3)^{2}}=\frac{A}{x+2}+\frac{B}{x-3}+\frac{c}{(x-3)^{2}}$
$2 x+1=A(x-3)^{2}+B(x+2)(x-3)+C(x+2)$
$2 x+1=A x^{2}-3 A x+9 A+B x^{2}-5 B x-6 B+C x+2 C$
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,
$I=\frac{11}{4} \int \frac{d x}{x+1}-\frac{5}{2} \int \frac{d x}{(x+1)^{2}}-\frac{11}{4} \int \frac{d x}{x+3}$
$I=\frac{11}{4} \log |x+1|-\frac{5}{2(x+1)}-\frac{11}{4} \log |x+3|+C$
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