Evaluate the following integral:
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 $\mathrm{x}$

$20=2 A+B+C$

$20=12-1+C$

$C=9$

$I=\int \frac{6 d x}{x}-\int \frac{d x}{x+1}+9 \int \frac{d x}{(x+1)^{2}}$

$I=6 \log |x|-\log |x+1|-\frac{9}{x+1}+C$