int frac{d x}{(x-1)^{frac{3}{4}}(x+2)^{frac{5 | vedclass

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Question

(A) Let $I = \int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$.We can rewrite the integrand as:$I = \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\fr

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

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(A) Let $I = \int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$.We can rewrite the integrand as:$I = \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{3}{4}}(x+2)^{\frac{2}{4}}} d x = \int \left(\frac{x-1}{x+2}\right)^{-\frac{3}{4}} \cdot \frac{1}{(x+2)^2} d x$.Let $t = \frac{x-1}{x+2}$.Then,$dt = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2} d x = \frac{3}{(x+2)^2} d x$.Thus,$\frac{1}{(x+2)^2} d x = \frac{dt}{3}$.Substituting these into the integral:$I = \int t^{-\frac{3}{4}} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-\frac{3}{4}} dt$.$I = \frac{1}{3} \cdot \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + C = \frac{4}{3} t^{\frac{1}{4}} + C$.Substituting back $t = \frac{x-1}{x+2}$,we get:$I = \frac{4}{3} \left(\frac{x-1}{x+2}\right)^{\frac{1}{4}} + C$.

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

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