TITLE: Evaluate $\int \frac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$
QUESTION [13 upvotes]: $$\int \dfrac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$$
Hey there, I've got this complicated integral to evaluate, but I don't know how to go about. I have tried making two substitutions:

$ t^2 = 1 - x $
$ x = \sin^2\theta $

But both gave another complicated integral to evaluate:
$$ \int \dfrac {\operatorname d\!t} {(t^2-1)\sqrt{ t^2 + t + 1 }} $$
I tried to get the answer for this one using wolfram alpha, but it gave a HUGE, simply HUGE solution. I also tried to get the solution for the original question via wolfram alpha, but it timed out.
Any ideas?

REPLY [3 votes]: The following integral
$$ \int \frac{1}{x \sqrt{ax^2 + bx + c}}\text{d}x$$
can be solved by substituting $x = \frac{1}{t}$ to get
$$\int \frac{\pm 1}{\sqrt{a + bt + ct^2}}\text{d}t$$
Which can be recast as the derivative of an inverse trigonometric function (could be hyperbolic, depending on the signs taken):
$$\int \frac{\pm 1}{\sqrt{\pm 1 \pm x^2}}$$
And as noted by Ron, your integral is the sum of two such integrals.