Computing $\int_{0}^{+\infty}\frac{\log(x)}{\sqrt x(1+{x^2})}dx$. I would like to compute the following integral :
$$\int_{0}^{+\infty}\frac{\log(x)}{\sqrt x(1+{x^2})}dx$$ using Residue theorem.
I took the contour corresponding to half of the "donuts" $\{r<|z|<R\}$.
So by theorem and an classical argument we have 
$I=2i\pi Res(f,i)$. 
My problem here, is that I need ton compute the residue for the principal argument of $\log(z)$ and $\sqrt{z}$. 
How can I do that ?
Note Bene, apologies for the 'unclear' question but I am on my mobile.
 A: Sub $x=y^2$ and the integral becomes
$$4 \int_0^{\infty} dy \frac{\log{y}}{1+y^4} $$
So consider
$$\oint_C dz \frac{\log^2{z}}{1+z^4} $$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis.  In the limit as $R \to \infty$ and $\epsilon \to 0$, the contour integral is equal to
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} \frac{dx}{1+x^4} $$
By the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand, or
$$i 2 \pi \left[\frac{(i \pi/4)^2}{4 e^{i 3 \pi/4}} +\frac{(i 3\pi/4)^2}{4 e^{i 9 \pi/4}}+\frac{(i 5 \pi/4)^2}{4 e^{i 15 \pi/4}}+\frac{(i 7 \pi/4)^2}{4 e^{i 21 \pi/4}}\right ] = \frac{\sqrt{2} \pi^3}{4} (4+i)$$
Now, by the residue theorem again, we may find similarly that**
$$-i 2 \pi \int_0^{\infty} \frac{dx}{1+x^4} = i 2 \pi \left[\frac{i \pi/4}{4 e^{i 3 \pi/4}} +\frac{i 3\pi/4}{4 e^{i 9 \pi/4}}+\frac{i 5 \pi/4}{4 e^{i 15 \pi/4}}+\frac{i 7 \pi/4}{4 e^{i 21 \pi/4}}\right ] = -i \frac{\sqrt{2} \pi^2}{2}$$
$$\implies \int_0^{\infty} \frac{dx}{1+x^4} = \frac{\sqrt{2} \pi}{4} $$
Putting all of the above results together, we find that
$$ \int_0^{\infty} dx \frac{\log{x}}{1+x^4} = -\frac{\sqrt{2} \pi^2}{16} $$
so that the original integral is

$$\int_0^{\infty} dx \frac{\log{x}}{\sqrt{x}(1+x^2)} = -\frac{\sqrt{2} \pi^2}{4} $$

** I know there are easier ways to get the value of that integral, but I prefer to be consistent with the method I use.
A: Hint: Let $u=\sqrt{x}$ to avoid the square root and then $x=u^2$ and the integral becomes
$$ I=4\int_0^\infty\frac{\log u}{1+u^4}du. $$
Then use the contour $\Gamma=\{z| r\le|z|\le R\}\setminus [r,R]$ to work on the function
$$ f(z)=\frac{\log^2 z}{1+z^4} $$
and you will get the anwser.
