Question

$\int \frac{x^2}{{\left(x-1\right)}^3\left(x+1\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{x^2}{{\left(x-1\right)}^3\left(x+1\right)}dx \\ \mathrm{Let}\frac{x^2}{{\left(x-1\right)}^3\left(x+1\right)}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{{\left(x-1\right)}^3}+\frac{D}{x+1}.....\left(1\right) \\ \Rightarrow x^2=A{\left(x-1\right)}^2\left(x+1\right)+B\left(x-1\right)\left(x+1\right)+C\left(x+1\right)+D{\left(x-1\right)}^3.....\left(2\right) \\ \mathrm{Putting}x=1\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=2C \\ \Rightarrow C=\frac{1}{2} \\ \mathrm{Putting}x=-1\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 1=-8D \\ \Rightarrow D=\frac{-1}{8} \end{gathered}$$

$$\begin{gathered} \mathrm{Putting}x=2\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 4=3A+3B+3C+D \\ \Rightarrow 4=3A+3B+\frac{3}{2}-\frac{1}{8} \\ \Rightarrow 3A+3B=4-\frac{3}{2}+\frac{1}{8} \\ \Rightarrow 3A+3B=\frac{32-12+1}{8} \\ \Rightarrow 3A+3B=\frac{21}{8} \\ \Rightarrow A+B=\frac{7}{8} \\ \mathrm{And}\mathrm{putting}x=0\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ 0=A-B+C-D \\ \Rightarrow 0=A-B+\frac{1}{2}+\frac{1}{8}\left[\because C=\frac{1}{2},D=\frac{1}{8}\right] \\ \Rightarrow A-B=-\frac{5}{8} \end{gathered}$$
$$\begin{gathered} \mathrm{Here},A+B=\frac{7}{8}\mathrm{and}A-B=-\frac{5}{8}\Rightarrow A=\frac{1}{8}\mathrm{and}B=\frac{3}{4} \\ \mathrm{Therefore},\left(1\right)\mathrm{becomes}, \\ \frac{x^2}{{\left(x-1\right)}^3\left(x+1\right)}=\frac{1}{8\left(x-1\right)}+\frac{3}{4{\left(x-1\right)}^2}+\frac{1}{2{\left(x-1\right)}^3}-\frac{1}{8\left(x+1\right)} \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes} \\ I=\int \left[\frac{1}{8\left(x-1\right)}+\frac{3}{4{\left(x-1\right)}^2}+\frac{1}{2{\left(x-1\right)}^3}-\frac{1}{8\left(x+1\right)}\right]dx \\ =\frac{1}{8}\log \left|x-1\right|-\frac{3}{4\left(x-1\right)}-\frac{1}{4{\left(x-1\right)}^2}-\frac{1}{8}\log \left|x+1\right|+C \\ =\frac{1}{8}\log \left|\frac{x-1}{x+1}\right|-\frac{3}{4\left(x-1\right)}-\frac{1}{4{\left(x-1\right)}^2}+C \end{gathered}$$