Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\left(x-1\right)\left(x+1\right)\left(x+2\right)} \\ \mathrm{Let}\frac{1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2} \\ \Rightarrow \frac{1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\frac{A\left(x+1\right)\left(x+2\right)+B\left(x-1\right)\left(x+2\right)+C\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)} \\ \Rightarrow 1=A\left(x+1\right)\left(x+2\right)+B\left(x-1\right)\left(x+2\right)+C\left(x-1\right)\left(x+1\right) \\ \mathrm{Putting}x-1=0 \\ \Rightarrow x=1 \\ 1=A\left(1+1\right)\left(1+2\right)+B\times 0+C\times 0 \\ \Rightarrow 1=A\times 6 \\ \Rightarrow A=\frac{1}{6} \\ \mathrm{Putting}x+1=0 \\ \Rightarrow x=-1 \\ 1=A\times 0+B\left(-2\right)\left(1\right)+C\times 0 \\ \Rightarrow B=-\frac{1}{2} \\ \mathrm{Putting}x+2=0 \\ \Rightarrow x=-2 \\ 1=A\times 0+B\times 0+C\left(-2-1\right)\left(-2+1\right) \\ \Rightarrow 1=C\times 3 \\ \Rightarrow C=\frac{1}{3} \\ \therefore I=\frac{1}{6}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{3}\int \frac{dx}{x+2} \\ =\frac{1}{6}\log \left|x-1\right|-\frac{1}{2}\log \left|x+1\right|+\frac{1}{3}\log \left|x+2\right|+C \\ =\frac{1}{6}\log \left|x-1\right|-\frac{3}{6}\log \left|x+1\right|+\frac{2}{6}\log \left|x+2\right|+C \\ =\frac{1}{6}\left[\log \left|x-1\right|-3\log \left|x+1\right|+2\log \left|x+2\right|\right]+C \\ =\frac{1}{6}\log \left|\frac{\left(x-1\right){\left(x+2\right)}^2}{{\left(x+1\right)}^3}\right|+C \end{gathered}$$