Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{xdx}{{\left(x-1\right)}^2\left(x+2\right)} \\ \mathrm{Let}\frac{x}{{\left(x-1\right)}^2\left(x+2\right)}=\frac{A}{x-1}+\frac{B}{{\left(x-1\right)}^2}+\frac{C}{x+2} \\ \Rightarrow \frac{x}{{\left(x-1\right)}^2\left(x+2\right)}=\frac{A\left(x-1\right)\left(x+2\right)+B\left(x+2\right)+C{\left(x-1\right)}^2}{{\left(x-1\right)}^2\left(x+2\right)} \\ \Rightarrow x=A\left(x^2+2x-x-2\right)+B\left(x+2\right)+C\left(x^2-2x+1\right) \\ \Rightarrow x=A\left(x^2+x-2\right)+B\left(x+2\right)+C\left(x^2-2x+1\right) \\ \Rightarrow x=\left(A+C\right)x^2+\left(A+B-2C\right)x+\left(-2A+2B+C\right) \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+C=0.....\left(1\right) \\ A+B-2C=1.....\left(2\right) \\ -2A+2B+C=0.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=\frac{2}{9},B=\frac{1}{3}\mathrm{and}C=-\frac{2}{9} \\ \therefore \frac{x}{{\left(x-1\right)}^2\left(x+2\right)}=\frac{2}{9\left(x-1\right)}+\frac{1}{3{\left(x-1\right)}^2}-\frac{2}{9\left(x+2\right)} \\ \Rightarrow I=\frac{2}{9}\int \frac{dx}{x-1}+\frac{1}{3}\int \frac{dx}{{\left(x-1\right)}^2}-\frac{2}{9}\int \frac{dx}{x+2} \\ =\frac{2}{9}\log \left|x-1\right|+\frac{1}{3}\times \left(\frac{-1}{x-1}\right)-\frac{2}{9}\log \left|x+2\right|+C \\ =\frac{2}{9}\log \left|\frac{x-1}{x+2}\right|-\frac{1}{3\left(x-1\right)}+C \end{gathered}$$