Question:
Evaluate the following integral:
$\int \frac{2 x^{2}+7 x-3}{x^{2}(2 x+1)} d x$
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,
$I=\int \frac{13 d x}{x}-\int \frac{3 d x}{x^{2}}-24 \int \frac{d x}{2 x+1}$
$I=13 \log |x|+\frac{3}{x}-12 \log |2 x+1|+C$
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