The integral int frac{1}{sqrt[4]{(x-1)^{3}(x+ | vedclass

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Question

(C) Let $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$.Rewrite the integrand as: $I = \int \frac{dx}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^{3/4

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

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(C) Let $I = \int \frac{dx}{(x-1)^{3/4}(x+2)^{5/4}}$.Rewrite the integrand as: $I = \int \frac{dx}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^{3/4} \cdot (x-1)^{5/4}} = \int \frac{dx}{\left(\frac{x+2}{x-1}\right)^{5/4} \cdot (x-1)^2}$.Let $t = \frac{x+2}{x-1}$. Then $dt = \frac{(x-1)(1) - (x+2)(1)}{(x-1)^2} dx = \frac{-3}{(x-1)^2} dx$,so $\frac{dx}{(x-1)^2} = -\frac{1}{3} dt$.Substituting these into the integral: $I = \int \frac{-1/3}{t^{5/4}} dt = -\frac{1}{3} \int t^{-5/4} dt$.Integrating: $I = -\frac{1}{3} \left( \frac{t^{-1/4}}{-1/4} \right) + C = \frac{4}{3} t^{-1/4} + C$.Substituting back $t = \frac{x+2}{x-1}$: $I = \frac{4}{3} \left( \frac{x+2}{x-1} \right)^{-1/4} + C = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C$.

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

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