int frac{d x}{(x-3)^{frac{4}{5}}(x+1)^{frac{6 | vedclass

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(A) Let $I = \int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}}$.We can rewrite the integral as:$I = \int \frac{d x}{(x+1)^2 \left(\frac{x-3}{x+1

Vedclass · Accepted Answer

$\frac{5}{4} \sqrt[5]{\frac{x-3}{x+1}} + C$

Vedclass · Answer

(A) Let $I = \int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}}$.We can rewrite the integral as:$I = \int \frac{d x}{(x+1)^2 \left(\frac{x-3}{x+1}\right)^{\frac{4}{5}}}$.Let $t = \frac{x-3}{x+1}$.Then,$dt = \frac{(x+1)(1) - (x-3)(1)}{(x+1)^2} dx = \frac{4}{(x+1)^2} dx$.Thus,$\frac{dx}{(x+1)^2} = \frac{dt}{4}$.Substituting these into the integral:$I = \int \frac{1}{t^{\frac{4}{5}}} \cdot \frac{dt}{4} = \frac{1}{4} \int t^{-\frac{4}{5}} dt$.$I = \frac{1}{4} \cdot \frac{t^{\frac{1}{5}}}{\frac{1}{5}} + C = \frac{5}{4} t^{\frac{1}{5}} + C$.Substituting back $t = \frac{x-3}{x+1}$:$I = \frac{5}{4} \left(\frac{x-3}{x+1}\right)^{\frac{1}{5}} + C$.

$\int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}} = $

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