# Integrating $\int \left(\sqrt[6]{\frac{x}{x-2}} - \sqrt[4]{\frac{x}{x-2}}\right)\frac{\mathrm dx}{x^2-2x}$ with partial fractions

Recently I've been studying on partial fractions and integration using partial fraction decomposition. I've not had any problems solving those types of integrals until I came across this integral:

$$\int \left(\sqrt[6]{\dfrac{x}{x-2}} - \sqrt[4]{\dfrac{x}{x-2}}\right)\frac{\mathrm dx}{x^2-2x}$$

The book hints that you should substitute $$\left( \dfrac{x}{x-2}=t^{12}, t \in \Bbb R \right)$$. I've tried countless times but haven't found any way as to even end up with an integrand that can be decomposed into partial fractions.

With you Substitution you will get $$x=\frac{-2t^{12}}{1-t^{12}}$$ and you can compute $$dx=…$$

$$\frac{x}{x-2}=t^{12}$$ $$d(\frac{x}{x-2})=\frac{-2}{(x-2)^2}dx=12t^{11}dt$$ $$\frac{dx}{x^2-2x}=\frac{-(x-2)^2}{2x(x-2)}(12t^{11})dt=\frac{-6t^{11}}{t^{12}}dt=-\frac{6}{t}dt$$ $$\int (\sqrt[6]{\frac{x}{x-2}} - \sqrt[4]{\frac{x}{x-2}})\frac{dx}{x^2-2x}=-6\int (t^2-t^3)\frac{1}{t}dt=\int (6t^2-6t)dt=2t^3-3t^2$$

Hint

For $$\dfrac x{x-2}\ge0,$$ we need either $$x>2$$ or $$x<0$$

So, WLOG $$x-1=\sec2t,dx=?$$

$$\dfrac x{x-2}=\dfrac{1+\sec2t}{\sec2t-1}=\tan^2t$$

$$x^2-2x=\tan^22t$$