If int frac{1}{x} sqrt{frac{1-sqrt{x}}{1+sqrt | vedclass

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(C) Let $I = \int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$. Substitute $\sqrt{x} = \cos 	heta$, so $x = \cos^2 	heta$ and $dx = -2 \cos

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$\log \left(\frac{1+\sqrt{1-x}}{\sqrt{x}}\right)$

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(C) Let $I = \int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$. Substitute $\sqrt{x} = \cos 	heta$, so $x = \cos^2 	heta$ and $dx = -2 \cos 	heta \sin 	heta d	heta$. Then $I = \int \frac{1}{\cos^2 	heta} \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} (-2 \cos 	heta \sin 	heta) d	heta$. Using $\sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = 	an(	heta/2)$, we get $I = \int \frac{1}{\cos^2 	heta} 	an(	heta/2) (-2 \cos 	heta \sin 	heta) d	heta = -2 \int \frac{\sin 	heta}{\cos 	heta} 	an(	heta/2) d	heta$. Since $\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$ and $\cos 	heta = \cos^2(	heta/2) - \sin^2(	heta/2)$, we have $I = -2 \int \frac{2 \sin(	heta/2) \cos(	heta/2)}{\cos^2(	heta/2) - \sin^2(	heta/2)} \frac{\sin(	heta/2)}{\cos(	heta/2)} d	heta = -4 \int \frac{\sin^2(	heta/2)}{\cos^2(	heta/2) - \sin^2(	heta/2)} d	heta$. Dividing numerator and denominator by $\cos^2(	heta/2)$, $I = -4 \int \frac{	an^2(	heta/2)}{1 - 	an^2(	heta/2)} d	heta = 4 \int \frac{1 - (1 - 	an^2(	heta/2))}{1 - 	an^2(	heta/2)} d	heta = 4 \int \left( \frac{1}{1 - 	an^2(	heta/2)} - 1 ight) d	heta$. This simplifies to $2 \log \left| \frac{1+	an(	heta/2)}{1-	an(	heta/2)} ight| - 2	heta + c$. Substituting $	an(	heta/2) = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$, we find $f(x) = \log \left( \frac{1+\sqrt{1-x}}{\sqrt{x}} ight)$.

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \operatorname{Sin}^{-1} \sqrt{x}+c$, then $f(x)=$

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