The integral int {frac{{dx}}{{{{(x + 1)}^{fra | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(x+1)^{3/4}(x-2)^{5/4}}$.We can rewrite the integral as:$I = \int \frac{dx}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^{3/4} (x-2)^{5/4}} = \int \frac{dx}{\left(\frac{x+1}{x-2}\right)^{3/4} (x-2)^2}$.Let $t = \frac{x+1}{x-2}$.Then,$dt = \frac{(x-2)(1) - (x+1)(1)}{(x-2)^2} dx = \frac{x-2-x-1}{(x-2)^2} dx = \frac{-3}{(x-2)^2} dx$.So,$\frac{dx}{(x-2)^2} = -\frac{1}{3} dt$.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 $t$:$I = -\frac{1}{3} \left( \frac{t^{-3/4 + 1}}{-3/4 + 1} \right) + c = -\frac{1}{3} \left( \frac{t^{1/4}}{1/4} \right) + c = -\frac{4}{3} t^{1/4} + c$.Substituting $t = \frac{x+1}{x-2}$ back:$I = -\frac{4}{3} \left( \frac{x+1}{x-2} \right)^{1/4} + c$.

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

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