Evaluate the following integrals:


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


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

By using partial differentiation,



By substituting the $x^{2}$ coefficients and other coefficients we can get,

$\mathrm{A}=-1 / 8 ; \mathrm{B}=1 / 8 ; \mathrm{C}=3 / 4 ; \mathrm{D}=1 / 2$

$=\int \frac{-d x}{8(x+1)}+\int \frac{d x}{8(x-1)}+\int \frac{3 d x}{4(x-1)^{2}}+\int \frac{d x}{2(x-1)^{3}}$

$=-\frac{1}{8} \log (1+x)+\frac{1}{8} \log (x-1)-\frac{3}{4(x-1)}-\frac{1}{4}\left(\frac{1}{1-x^{2}}\right)+c$

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