Question

$\int \frac{x^2+6x-8}{x^3-4x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^2+6x-8}{x^3-4x}\right)dx \\ =\int \frac{\left(x^2+6x-8\right)}{x\left(x^2-4\right)}dx \\ =\int \frac{\left(x^2+6x-8\right)}{x\left(x-2\right)\left(x+2\right)}dx \\ \mathrm{Let}\frac{x^2+6x-8}{x\left(x-2\right)\left(x+2\right)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2} \\ \Rightarrow \frac{x^2+6x-8}{x\left(x-2\right)\left(x+2\right)}=\frac{A\left(x-2\right)\left(x+2\right)+B\left(x\right)\left(x+2\right)+C\left(x\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)} \\ \Rightarrow x^2+6x-8=A\left(x^2-4\right)+B\left(x^2+2x\right)+C\left(x^2-2x\right) \\ \mathrm{Putting}x-2=0 \\ \Rightarrow x=2 \\ 4+6\times 2-8=A\times 0+B\left(4+4\right) \\ \Rightarrow 8=B\times 8 \\ \Rightarrow B=1 \\ \mathrm{Putting}x=-2 \\ 4-12-8=A\times 0+B\times 0+C\times 8 \\ \Rightarrow C=-2 \\ \mathrm{Putting}x=0 \\ -8=A\left(-4\right)+B\times 0+C\times 0 \\ \Rightarrow A=2 \\ \therefore I=\int \frac{2}{x}+\int \frac{dx}{x-2}-2\int \frac{dx}{x+2} \\ =2\log \left|x\right|+\log \left|x-2\right|-2\log \left|x+2\right|+C \\ =\log x^2+\log \left|x-2\right|-\log {\left|x+2\right|}^2+C \\ =\log \left|\frac{x^2\left(x-2\right)}{{\left(x+2\right)}^2}\right|+C \end{gathered}$$