I want to evaluate this integral below using contour integral. $$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}(1+x^2)}$$ I know this can be done transforming x to sin or cos, but I want to solve this by techniques of complex analysis.
I tried key hole contour. I understand that the residues at 1 and -1 can’t be calculated as usual. Therefore I excluded those branch points by key hole contour. There are two small circles whose centers are 1 and -1, and two way path connects them. Using the property of branch, twice of the desired integral can be gained from the two way path. In order to exclude 1 and -1, I added one large circle which surrounds all of them.
In the contour, there are residues of i and -i, but the sum is equal to 0. Because integrals relates to circles vanishes under proper limit, this results in the desired integral is equal to 0, which is not the answer.
What is my misunderstanding?
You should consider a contour integral as in fig \begin{gather} \oint \frac{dz}{g(z)(1+z^2)}=\int_{-1}^1\frac{dx}{g(x+i0)(1+x^2)}-\int_{-1}^1\frac{dx}{g(x-i0)(1+x^2)},\ \ \ g(z)=\sqrt{1-z^2}. \end{gather} Now you fixate a regular branch of your multivalued $g(z)$ as
\begin{gather} g(x+i0)>0,\ \ x\in(-1,1)\ \ {\rm fixation\ of\ a\ regular\ branch} \end{gather} That means that $g(x-i0)=-g(x+i0)$ and also, that means that your original integral $I$ is equal to the integral along the upper bank of the contour. This way: \begin{gather} \oint=2I. \end{gather} On the other hand:
\begin{gather} \oint=2\pi i\Big(\underset{z=i}{\rm res}\frac{1}{(1+z^2)g(z)}+\underset{z=-i}{\rm res}\frac{1}{(1+z^2)g(z)}\Big)=2\pi i\Big(\frac{1}{2i g(i)}+\frac{1}{-2i g(-i)}\Big). \end{gather} Now from the definition of your regular branch $g(i)=\sqrt{2}$ and $g(-i)=-\sqrt{2}$. Then: \begin{gather} I=\frac{1}{2}\oint=\frac{\pi}{\sqrt{2}}. \end{gather}