The integral dx[4](x-1)^3(x+2)^5 is equal to (where C is a constant of integration)

Question

The integral $\displaystyle\int\frac{dx}{\sqrt[4]{(x-1)^3(x+2)^5}}$ is equal to (where C is a constant of integration)

Accepted Answer

$$I=\int\frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}=\int\left(\frac{x+2}{x-1}\right)^{-5/4}\frac{dx}{(x-1)^2}$$

Let $\dfrac{x+2}{x-1}=t\Rightarrow\dfrac{(x-1)-(x+2)}{(x-1)^2}dx=dt\Rightarrow-\dfrac{3}{(x-1)^2}dx=dt$

$$I=-\frac13\int t^{-5/4}dt=\frac43\cdot t^{-1/4}+C=\frac43\left(\frac{x-1}{x+2}\right)^{1/4}+C$$

Solutions · 1