Evaluate the following integral:

Solution:

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

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

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

Put $x=1$

$1=4 \mathrm{~A}$

$A=\frac{1}{4}$

Put $x=-1$

$1=-2 C$

$C=-\frac{1}{2}$

Equating coefficients of $x^{2}$Equating coefficients of $x^{2}$

$1=A+B$

$1=\frac{1}{4}+B$

$B=\frac{3}{4}$

Thus,

$I=\frac{1}{4} \int \frac{d x}{x-1}+\frac{3}{4} \int \frac{d x}{x+1}-\frac{1}{2} \int \frac{d x}{(x+1)^{2}}$

$I=\frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C$