<s>[INST] Evaluation of $ \int\frac{1}{2x\sqrt{1-x}\cdot \sqrt{2-x+\sqrt{1-x}}}dx$ 
Evaluation of $\displaystyle \int\frac{1}{2x\sqrt{1-x}\cdot \sqrt{2-x+\sqrt{1-x}}}dx$

$\bf{My\; Try:}$ Let $x=\sin^2 \theta\;,$ Then $dx = 2\sin \theta \cdot \cos \theta d\theta$
So Integral $$\displaystyle I = \int\frac{2\sin \theta \cos \theta }{2\sin^2 \theta \cos \theta\sqrt{2-\sin^2 \theta+\cos \theta}}d\theta = \int\frac{1}{\sin \theta\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta$$
So $$\displaystyle I = \int\frac{\sin \theta}{(1-\cos^2 \theta)\sqrt{\cos^2 \theta+\cos \theta+1}}d\theta\;,$$ Now put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$. So $$\displaystyle I = \int\frac{1}{(1-t^2)\sqrt{t^2+t+1}}dt$$
How can we continue?
 [/INST] I give you a step or two, and know you can finish from them:
Make a partial fraction decomposition of the rational part, and write your integral as
$$
\frac{1}{2}\int\frac{1}{(1-t)\sqrt{1+t+t^2}}\,dt+\frac{1}{2}\int\frac{1}{(1+t)\sqrt{1+t+t^2}}\,dt
$$
In the first integral, let
$$
u=\frac{1-t}{\sqrt{1+t+t^2}},
$$
and in the second, let
$$
u=\frac{1+t}{\sqrt{1+t+t^2}}.
$$
You will end up with integrals of the form
$$
\int\frac{1}{c-u^2}\,du
$$
which I'm certain you can handle.
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