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

We have to evaluate the integral

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

The radical $$\sqrt{x-x^{2}}$$ suggests writing the product $$x(1-x)$$ inside a single square root. A standard trigonometric substitution for this form is to let $$x=\sin^{2}\theta\;,$$ because $$1-\sin^{2}\theta=\cos^{2}\theta$$ turns the whole product into a perfect square.

So we set $$x=\sin^{2}\theta\;.$$ Then $$\sqrt{x}=\sin\theta$$ and the differential changes according to the rule $$dx=\frac{d}{d\theta}(\sin^{2}\theta)\,d\theta=2\sin\theta\cos\theta\,d\theta\;.$$

Now we rewrite every piece of the integrand:

$$\sqrt{x-x^{2}} =\sqrt{\sin^{2}\theta(1-\sin^{2}\theta)} =\sqrt{\sin^{2}\theta\cos^{2}\theta} =\sin\theta\cos\theta\;.$$

Substituting these expressions into the integral gives

$$I=\int\frac{2\sin\theta\cos\theta\,d\theta} {(1+\sin\theta)\;(\sin\theta\cos\theta)} =\int\frac{2\,d\theta}{1+\sin\theta}\;.$$

Thus the original integral has been reduced to

$$I=2\int\frac{d\theta}{1+\sin\theta}\;.$$

To integrate $$\displaystyle\frac{1}{1+\sin\theta},$$ we multiply numerator and denominator by its conjugate $$1-\sin\theta$$ so that a Pythagorean identity can be used:

$$\frac{1}{1+\sin\theta} =\frac{1-\sin\theta}{(1+\sin\theta)(1-\sin\theta)} =\frac{1-\sin\theta}{1-\sin^{2}\theta} =\frac{1-\sin\theta}{\cos^{2}\theta}\;,$$

because $$1-\sin^{2}\theta=\cos^{2}\theta.$$

Therefore

$$I=2\int\Bigg(\frac{1}{\cos^{2}\theta}-\frac{\sin\theta}{\cos^{2}\theta}\Bigg)\,d\theta =2\int\bigl(\sec^{2}\theta-\sin\theta\sec^{2}\theta\bigr)\,d\theta\;.$$

We split the integral term by term:

$$I=2\Bigl[\int\sec^{2}\theta\,d\theta -\int\sin\theta\,\sec^{2}\theta\,d\theta\Bigr]\;.$$

The standard antiderivatives are

$$\int\sec^{2}\theta\,d\theta=\tan\theta\quad\text{and}\quad \int\sin\theta\sec^{2}\theta\,d\theta =\int\sin\theta\frac{1}{\cos^{2}\theta}\,d\theta =\int\frac{\sin\theta}{\cos^{2}\theta}\,d\theta =\sec\theta\;,$$

because the derivative of $$\sec\theta$$ is $$\sec\theta\tan\theta,$$ and integrating $$\tan\theta\sec\theta$$ exactly reproduces $$\sec\theta.$$ Hence

$$I=2\bigl(\tan\theta-\sec\theta\bigr)+C\;.$$

We must now return to the original variable $$x.$$ From the substitution $$x=\sin^{2}\theta,$$ we have

$$\sin\theta=\sqrt{x},\qquad \cos\theta=\sqrt{1-\sin^{2}\theta}=\sqrt{1-x}\;.$$

Therefore

$$\tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{\sqrt{x}}{\sqrt{1-x}} =\sqrt{\frac{x}{1-x}},$$

and

$$\sec\theta=\frac{1}{\cos\theta} =\frac{1}{\sqrt{1-x}}\;.$$

Substituting these back gives

$$I=2\Bigl(\sqrt{\frac{x}{1-x}}-\frac{1}{\sqrt{1-x}}\Bigr)+C =\frac{2\bigl(\sqrt{x}-1\bigr)}{\sqrt{1-x}}+C\;.$$

We now notice that $$1-x=(1-\sqrt{x})(1+\sqrt{x}).$$ Using this factorisation we rewrite $$I$$ in a more elegant radical form:

$$\frac{2(\sqrt{x}-1)}{\sqrt{(1-\sqrt{x})(1+\sqrt{x})}} =\frac{-2(1-\sqrt{x})}{\sqrt{(1-\sqrt{x})(1+\sqrt{x})}} =-2\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}} =-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\;.$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.