The value of int frac{d x}{(x+1)^{3 / 4}(x-2) | vedclass

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

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$\frac{-4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+c$,where $c$ is a constant of integration.

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(D) Let $I = \int \frac{d x}{(x+1)^{3/4} (x-2)^{5/4}}$.We can rewrite the integrand as:$I = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{3/4} (x-2)^{5/4}} = \int \frac{d x}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^2}$.Let $t = \frac{x+1}{x-2}$.Then,$dt = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} dx = \frac{-3}{(x-2)^2} dx$.Thus,$\frac{dx}{(x-2)^2} = -\frac{1}{3} dt$.Substituting these into the integral:$I = \int \frac{1}{t^{3/4}} \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int t^{-3/4} dt$.Integrating with respect to $t$:$I = -\frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + c = -\frac{4}{3} t^{1/4} + c$.Substituting back $t = \frac{x+1}{x-2}$:$I = -\frac{4}{3} \left(\frac{x+1}{x-2}\right)^{1/4} + c$.

The value of $\int \frac{d x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to

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