Question

$\int \frac{x^2-2}{x^5-x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^2-2}{x^5-x}\right)dx \\ =\int \frac{\left(x^2-2\right)}{x\left(x^4-1\right)}dx \\ =\int \frac{x\left(x^2-2\right)}{x^2\left(x^2-1\right)\left(x^2+1\right)}dx \\ \mathrm{Putting}{x}^2=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \frac{\left(t-2\right)}{t\left(t-1\right)\left(t+1\right)}dt \\ \mathrm{Let}\frac{t-2}{t\left(t-1\right)\left(t+1\right)}=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+1} \\ \Rightarrow \frac{t-2}{t\left(t-1\right)\left(t+1\right)}=\frac{A\left(t-1\right)\left(t+1\right)+Bt\left(t+1\right)+Ct·\left(t-1\right)}{t\left(t-1\right)\left(t+1\right)} \\ \Rightarrow t-2=A\left(t-1\right)\left(t+1\right)+Bt\left(t+1\right)+Ct\left(t-1\right) \\ \mathrm{Putting}t=1 \\ \therefore 1-2=B\times 2 \\ \Rightarrow B=-\frac{1}{2} \\ \mathrm{Putting}t=0 \\ \therefore -2=A\left(-1\right) \\ \Rightarrow A=2 \\ \mathrm{Putting}t=-1 \\ \therefore -3=C\left(-1\right)\left(-2\right) \\ \Rightarrow C=-\frac{3}{2} \\ \therefore I=\frac{2}{2}\int \frac{dt}{t}-\frac{1}{2\times 2}\int \frac{dt}{t-1}-\frac{3}{2\times 2}\int \frac{d}{t+1} \\ =\log \left|t\right|-\frac{1}{4}\log \left|t-1\right|-\frac{3}{4}\log \left|t+1\right|+C \\ =\log \left|x^2\right|-\frac{1}{4}\log \left|x^2-1\right|-\frac{3}{4}\log \left|x^2+1\right|+C \\ =2\log \left|x\right|-\frac{1}{4}\log \left|x^2-1\right|-\frac{3}{4}\log \left|x^2+1\right|+C \end{gathered}$$