I am having trouble with this integral from the 2012 MIT Integration Bee: $\int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$

Question

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

Could someone explain to me how to integrate this integral.

Thank you and cheers.

Answer 1

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

Let's put $x=\sin^{2}t$, then \begin{align*} \int\frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}&=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t-\sin^4 t}}} \\ &=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sqrt{\sin^2 t(1-\sin^2 t)}}}\\ &=\int{\frac{2\sin t\cos t dt}{(1+\sin t)\sin t\cos t}}\\ &=\int{\frac{2dt}{1+\sin t}}\\ &=\int{\frac{2(1-\sin t)dt}{\cos^2 t}}=2\int{\sec^2 tdt}-2\int{\sec t\tan tdt}\\ &=2\tan t - 2\sec t + C\\ &=\frac{2(\sin t - 1)}{\cos t} + C\\ &=\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+C \end{align*}