int frac{d x}{(x-1) sqrt{x+2}} = | vedclass

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(D) Let $I = \int \frac{d x}{(x-1) \sqrt{x+2}}$.Substitute $t = \sqrt{x+2}$,which implies $t^2 = x+2$,so $x = t^2 - 2$.Then $dx = 2t \, dt$.Substituti

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+C$

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(D) Let $I = \int \frac{d x}{(x-1) \sqrt{x+2}}$.Substitute $t = \sqrt{x+2}$,which implies $t^2 = x+2$,so $x = t^2 - 2$.Then $dx = 2t \, dt$.Substituting these into the integral:$I = \int \frac{2t \, dt}{(t^2 - 2 - 1)t} = \int \frac{2 \, dt}{t^2 - 3}$.Using the standard integral formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} ight| + C$:$I = 2 	imes \frac{1}{2\sqrt{3}} \log \left| \frac{t-\sqrt{3}}{t+\sqrt{3}} ight| + C$.$I = \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}} ight| + C$.

$\int \frac{d x}{(x-1) \sqrt{x+2}} = $

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