$\int _0 ^\infty \frac{1}{(x+1)(1-x) \sqrt{x}} dx$ How do I have to compute the integral $\int _0 ^\infty \frac{1}{(x+1)(1-x) \sqrt{x}} dx$? I think maybe the residue theorem is useful, but there are poles at $0$ and $1$, so I don't see what do I have to do.. 
 A: To elaborate on what @xpaul mentioned in the comments,
You can use the substitution, $$x^{1/2} = u$$
Differentiating,
$$2du = \frac{dx}{x^{1/2}}$$
Therefore the integral reduces to 
$$\int_{0}^{\infty}\frac{2du}{(1-u^{2})({1+u^{2})}}$$
Splitting the integral using partial fractions,
$$-\int_{0}^{\infty}\frac{du}{u^{2}+1}+\int_{0}^{\infty}\frac{du}{u^{2}-1}$$
The first integral equals $\pi/2$ while the second one diverges
A: As noted, the integral $\int_0^\infty$ diverges because of the pole at $x=1$.   
The indefinite integral is:
$$
\int \!{\frac {1}{ \left( 1-x \right)  \left( 1+x \right) \sqrt {x}}}
\,{\rm d}x=-\frac{\ln  \left( -1+\sqrt {x} \right)}{2} +\arctan \left( 
\sqrt {x} \right) +\frac{\ln  \left( 1+\sqrt {x} \right) }{2}+ C
$$
(of course possibly with different $C$ on opposite sides of $x=1$).
We may try for a "principal value" like this.
$$
\lim_{a\searrow 0}\left(\int_0^{1-a}\frac{dx}{(1-x)(1+x)\sqrt{x}\;}
+\int_{1+a}^{+\infty}\frac{dx}{(1-x)(1+x)\sqrt{x}\;}\right) = \frac{\pi}{2}
$$
