Evaluate the following integral:

Solution:

$I=\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x$

$\frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{2 x+1}$

$2 x^{2}+7 x-3=A x(2 x+1)+B(2 x+1)+C x^{2}$

Equating constants

$-3=B$

Equating coefficients of $x$

$7=A+2 B$

$7=A-6$

$A=13$

Equating coefficients of $x^{2}$

$2=2 A+C$

$2=26+C$

$C=-24$

Thus,

$I=\int \frac{13 d x}{x}-\int \frac{3 d x}{x^{2}}-24 \int \frac{d x}{2 x+1}$

$I=13 \log |x|+\frac{3}{x}-12 \log |2 x+1|+C$