$∫\frac{dx}{(1+\sqrt{x})\sqrt{(x-x^2)}}$ is equal to

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

Let $I=∫\frac{dx}{(1+\sqrt{x})\sqrt{(x-x^2)}}$.

If $\sqrt{x}=sin\,p$, then $\frac{1}{2\sqrt{x}}dx=cos\,p\,dp$

$⇒I=∫\frac{2sin\,p\,cos\,p\,dp}{(1+sin\,p)sin\,p\,cos\,p}=2∫\frac{dp}{(1+sin\,p)}=2∫\frac{(1-sin\,p)dp}{cos^2p}$

$=2\{∫sec^2pdp-∫(tan\,p\,sec\,p)dp\}$

= 2 (tan p - sec p) + c = 2$(\sqrt{\frac{x}{(1-x)}}-\frac{1}{\sqrt{(1-x)}})+c=\frac{2(\sqrt{x}-1)}{\sqrt{(1-x)}}+c$

Hence (D) is the correct answer.

Alternatively: Write x = t² and then 1 + t = 1/k.