Question:
Evaluate $\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$
Solution:
$=\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$\
By using partial differentiation,
$=\frac{x^{2}}{(x-1)^{3}(x+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(x-1)^{2}}+\frac{D}{(x-1)^{3}}$
$x^{2}=A(x-1)^{3}+B(x-1)^{2}(x+1)+C(x-1)^{1}(x+1)+D(x+1)$
By substituting the $x^{2}$ coefficients and other coefficients we can get,
$\mathrm{A}=-1 / 8 ; \mathrm{B}=1 / 8 ; \mathrm{C}=3 / 4 ; \mathrm{D}=1 / 2$
$=\int \frac{-d x}{8(x+1)}+\int \frac{d x}{8(x-1)}+\int \frac{3 d x}{4(x-1)^{2}}+\int \frac{d x}{2(x-1)^{3}}$
$=-\frac{1}{8} \log (1+x)+\frac{1}{8} \log (x-1)-\frac{3}{4(x-1)}-\frac{1}{4}\left(\frac{1}{1-x^{2}}\right)+c$