Finding the correct contour to evaluate an integral with finite bounds of integration What I am interested in is finding a closed form solution to the following integral:
$$\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$$
My approach so far is as follows:
Let us consider $\Gamma = \gamma_1+\gamma_2$ where $\gamma_1$ is the path defined on the real axis from $-1$ to $1$. $\gamma_2$ is defined as $\textbf{not sure...}$. Thus, by the Residual Theorem, (under the assumption that our contour is in the upper half plane), we have:
$$\int_{\Gamma}f=2\pi i \text{Res}(f;i)=2\pi i \bigg( \frac{\sqrt{2}}{2i}\bigg)=\pi \sqrt{2}$$
And with the help of Mathematica, one finds that:
$$\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx=\pi(\sqrt{2}-1)$$
Naturally, I'm simply stuck on defining the contour to yield a nice integral to which should evaluate to $\pi$. From my understanding, defining $\gamma_2 = e^{it}, t\in [0, \pi]$ will not work since $i$ lies on such a contour.
 A: 
If one wishes to use complex analysis to evaluate the integral, $\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2}\,dx$, then one can proceed as follows. 

Let $f(z)=\frac{\sqrt{1-z^2}}{1+z^2}$. Analyze the contour integral
$$I=\oint_C \frac{\sqrt{1-z^2}}{1+z^2}\,dz$$
where $C$ is the classical "dog-bone" or "dumbbell" contour.  
Then, accounting for the residues from the poles at $z=\pm i$ and the Residue at Infinity we have
$$
\begin{align}
I&=2\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}\,dx\\\\
&=2\pi i \text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2}, z=\pm i,\infty\right)\\\\
&=2\pi i  \left(\frac{\sqrt 2}{2i}+\frac{-\sqrt 2}{-2i}+i\right)\\\\
&=2\pi \left(\sqrt 2 -1\right)
\end{align}$$
whereupon dividing by $2$ yields the coveted integral
$$\int_{-1}^1 \frac{\sqrt{1-z^2}}{1+z^2}\,dz=\pi(\sqrt 2 -1)$$
as was to be shown!

Instead of appealing to the residue at infinity, we can alternatively and equivalently  analyze the integral of $\frac{\sqrt{1-z^2}}{1+z^2}$ around $C$, where $C$ is a circle of radius $R$, centered at the origin and let $R\to \infty$.  Then, we have
$$\begin{align}
\lim_{R\to \infty}\oint_{|z|=R}\frac{\sqrt{1-z^2}}{1+z^2}\,dz&=\lim_{R\to \infty}\int_0^{2\pi}\frac{\sqrt{1-R^2e^{i2\phi}}}{1+R^2e^{i\phi}}\,iRe^{i\phi}\,d\phi\\\\
&=2\pi\\\\
&=2\pi i \text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2},z=\pm i\right)-2\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}\,dx
\end{align}$$
whereupon solving for the integral of interest yields the expected result!
A: I would like to add some  commentary to the excellent answer by @DrMV,
showing how to compute the residues involved. We will use
$$f(z) = \frac{1}{1+z^2}
\exp(1/2 \times\mathrm{LogA}(1+z)) 
\exp(1/2 \times\mathrm{LogB}(1-z)).$$
Here $\mathrm{LogA}$  denotes the branch of the  logarithm where $-\pi
\lt \arg \mathrm{LogA} \le \pi$  and $\mathrm{LogB}$ where $0 \lt \arg
\mathrm{LogB} \le 2\pi.$ The branch cut from $\mathrm{LogB}$ is inside
the dogbone  contour while  for $x\lt -1$  both branch cuts  apply. In
fact they cancel and we have  continuity across the cut and may derive
analyticity   by  Morera's   theorem   as  explained   at  this   MSE
link.
For the continuity the rational  factor is obviously the same above
and below  the cut, while for  the two logarithmic factors  we get for
$x\lt -1$ and above the  cut $\mathrm{LogA}(1+x) = \log(-x-1) + \pi i$
and $\mathrm{LogB}(1-x)  = \log(-x+1) +  2\pi i$ (rotation)  and below
the   cut   $\mathrm{LogA}(1+x)   =    \log(-x-1)   -   \pi   i$   and
$\mathrm{LogB}(1-x) = \log(-x+1)$  (rotation again). This yields above
the cut
$$\exp(1/2\times(\log(-x-1)+\pi i))\exp(1/2\times(\log(-x+1)+2\pi i))
\\ = \sqrt{x^2-1} \exp(3/2 \times \pi i) = -i\sqrt{x^2-1}$$
and below the cut
$$\exp(1/2\times(\log(-x-1)-\pi i))\exp(1/2\times(\log(-x+1)))
\\ = \sqrt{x^2-1} \exp(-1/2 \times \pi i) = -i\sqrt{x^2-1}$$
and we have continuity across the  cut. (For the cut itself the values
are  from the  above-the-cut case.)  We need  to verify  that  the two
segments above and below the single branch cut enclosed by the dogbone
contour are multiples of the target integral. We get above the cut
$$\exp(1/2\times(\log(x+1))\exp(1/2\times(\log(-x+1)+2\pi i))
= - \sqrt{1-x^2}$$
and below
$$\exp(1/2\times(\log(x+1))\exp(1/2\times(\log(-x+1)))
= \sqrt{1-x^2}.$$
This means that  with a counter-clockwise traversal of  the contour we
pick up twice the target integral. Next to compute the residues we get
for the easy ones at $\pm i$
$$\mathrm{Res}_{z=i} f(z) \\ =
\frac{1}{2i}\exp(1/2\times (\log \sqrt{2} + i\pi/4))
\exp(1/2\times (\log \sqrt{2} + (2\pi i-i\pi/4)))
\\ = -\frac{1}{2i} \sqrt{2} = \frac{i\sqrt{2}}{2}$$
and
$$\mathrm{Res}_{z=-i} f(z) \\ =
-\frac{1}{2i}\exp(1/2\times (\log \sqrt{2} - i\pi/4))
\exp(1/2\times (\log \sqrt{2} + i\pi/4))
\\ = -\frac{1}{2i} \sqrt{2} = \frac{i\sqrt{2}}{2}.$$
For  the residue  at infinity  we use  (no branch  cut  anymore around
infinity)
$$\mathrm{Res}_{z=\infty} f(z) =
- \lim_{R\rightarrow\infty} \frac{1}{2\pi i} 
\int_{|z|=R} f(z) \; dz.$$
Putting $z= R \exp(i\theta)$ we obtain
$$\int_0^{2\pi} \frac{1}{1+R^2 \exp(2i\theta)}
\exp(1/2\times\mathrm{LogA}(1+R\exp(i\theta))) \\ \times
\exp(1/2\times\mathrm{LogB}(1-R\exp(i\theta)))
Ri\exp(i\theta)d\theta.$$
We have two  intervals, from $0$ to $\pi$ (upper  half plane) and from
$\pi$ to $2\pi$  (lower half plane).  In the upper  half plane we have
as $R$ goes to infinity that
$$\mathrm{LogA}(1+R\exp(i\theta)) 
\rightarrow \log(R) + i\theta$$
and
$$\mathrm{LogB}(1-R\exp(i\theta)) 
\rightarrow \log(R) + i(\theta+\pi)$$
We obtain for the limit
$$\int_0^{\pi} \frac{R}{\exp(-2i\theta)+R^2}
\exp(\log R) \exp(i\theta + \pi i/2) i \exp(-i\theta)\; d\theta
\\ = - \int_0^{\pi} \frac{R^2}{\exp(-2i\theta)+R^2} \; d\theta
\rightarrow -\pi.$$
In the lower half plane we get
$$\mathrm{LogA}(1+R\exp(i\theta)) 
\rightarrow \log(R) + i(\theta-2\pi)$$
and
$$\mathrm{LogB}(1-R\exp(i\theta)) 
\rightarrow \log(R) + i(\theta-\pi)$$
This is the  sames as before except $\exp(\pi  i/2)$ has been replaced
by $\exp(-3\pi i/2)$ which makes  no difference (both evaluate to $i$)
and we once more obtain $-\pi$, for a total residue of $-(-2\pi)/(2\pi
i) = -i.$ 
Now the contour produces twice the desired value as explained earlier
and hence it is given by (poles outside rather than inside contour)
$$\frac{1}{2} \times - 2\pi i \times
\left(-i + i\sqrt{2}\right)$$
which is
$$\bbox[5px,border:2px solid #00A000]{
\pi(\sqrt{2}-1)}$$
as claimed.
