Evaluate $\int \frac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$

Solution 1:

For what it's worth, you can break things up a bit by recognizing that

$$\frac{1}{t^2-1} = \frac12 \left ( \frac{1}{t-1} - \frac{1}{t+1}\right )$$

For example, consider the $t-1$ piece; you may substitute $u=t+1/2$ and get

$$\int \frac{dt}{(t-1) \sqrt{t^2+t+1}} = \int \frac{du}{(u-3/2) \sqrt{u^2+3/4}}$$

This latter integral is relatively tame according to WA:

$$ \frac{1}{\sqrt{3}} \left[\log{\left(u-\frac{3}{2}\right)}-\log{\left(\sqrt{12 u^2+9}+3 u+\frac{3}{2}\right)}\right]+C$$

where $C$ is a constant of integration. A similar expression may be found for the $t+1$ piece.

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.