Question:
Evaluate the following integral:
$\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x} d x$
Solution:
$I=\int \frac{5 x^{2}+20 x+6}{x^{3}+2 x^{2}+x}=\int \frac{5 x^{2}+20 x+6}{x(x+1)^{2}}$
$\frac{5 x^{2}+20 x+6}{x(x+1)^{2}}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$
$5 x^{2}+20 x+6=A(x+1)^{2}+B x(x+1)+C x$
Equating constants
$6=A$
Equating coefficients of $x^{2}$
$5=A+B$
$B=-1$
Equating coefficients of $\mathrm{x}$
$20=2 A+B+C$
$20=12-1+C$
$C=9$
$I=\int \frac{6 d x}{x}-\int \frac{d x}{x+1}+9 \int \frac{d x}{(x+1)^{2}}$
$I=6 \log |x|-\log |x+1|-\frac{9}{x+1}+C$