The integral
Question:

The integral $\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} \mathrm{dx}$ is equal to :

(where $C$ is a constant of integration)

1. $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C$

2. $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C$

3. $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

4. $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C$

Correct Option: , 3

Solution:

$\int \frac{d x}{(x-1)^{3 / 4}(x+2)^{5 / 4}}$

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

put $\frac{x+2}{x-1}=t$

$=-\frac{1}{3} \int \frac{\mathrm{dt}}{\mathrm{t}^{5 / 4}}$

$=\frac{4}{3} \cdot \frac{1}{\mathrm{t}^{1 / 4}}+\mathrm{C}$

$=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1 / 4}+C$