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

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(C) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.Substitute $\sqrt{x} = t$,then $\frac{1}{2

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$-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$

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(C) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.Substitute $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2dt$.The integral becomes $I = \int \frac{2dt}{(1+t) \sqrt{1-t^2}}$.Further,substitute $t = \sin 	heta$,then $dt = \cos 	heta d	heta$.$I = \int \frac{2 \cos 	heta d	heta}{(1+\sin 	heta) \sqrt{1-\sin^2 	heta}} = \int \frac{2 \cos 	heta d	heta}{(1+\sin 	heta) \cos 	heta} = 2 \int \frac{d	heta}{1+\sin 	heta}$.Using $1+\sin 	heta = 1+\cos(\frac{\pi}{2}-	heta) = 2 \cos^2(\frac{\pi}{4}-\frac{	heta}{2})$,we get $I = \int \sec^2(\frac{\pi}{4}-\frac{	heta}{2}) d	heta$.Integrating,$I = -2 	an(\frac{\pi}{4}-\frac{	heta}{2}) + c = -2 \frac{1-	an(	heta/2)}{1+	an(	heta/2)} + c$.Alternatively,using the substitution $u = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$,we find $I = -2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} + c$.

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

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