Integrate $\frac{\sqrt{1-x^2}}{1+x^2}$ from $-1$ to $1$ using the residue theorem

Question

I tried integrating using the "dog bone", (wich idea is explained fairly well in this post: Integrating around a dog bone contour). I oriented the curve counter-clockwise.

Integrating along the dog bone (and computing the limit) gives me $-2I$, where $I$ is the integral i have to compute. Then computing the residues I find that the residues at $i$ and $-i$ are opposite, so they cancel each other out, while the one at infinity equals $-i$. Using the residue theorem, I find that $-2I=-2i\pi*(-i)$, but that is different from $(\sqrt2 - 1)*\pi$ which is what Wolfram Alpha gives me.

What did I do wrong?

I calculated the limit, for z approaching the point i or -i, of f(z)(z-i) or f(z)(z+i) — Omega, May 12 '19 at 00:10
For what it's worth, I think you've asked your question well. You could possibly include a few more details and present it slightly better using more MathJax, but I think your post is otherwise quite good. — Clayton, May 12 '19 at 00:10
For the one at infinity, I calculated the residue in 0 of $g(z)=-\frac{1}{z^2}*f(\frac{1}{z})$ — Omega, May 12 '19 at 00:12
@Omega Please outline your calculations more. See https://math.stackexchange.com/questions/2216820/computing-int-11-frac-sqrt1-x2x21-using-residue-calculus?noredirect=1&lq=1 for a complete proof. Note your residues should not have opposite sign if you are keeping track of branch cuts properly; best to not trust mathematica on this one — Brevan Ellefsen, May 12 '19 at 08:43
You thus get $$\left(-\tfrac{i\sqrt{2}}{2} -\tfrac{i\sqrt{2}}{2} \right) \cdot (-2 \pi i) + 2I = -2\pi i \cdot (-i)$$ i.e. that $$2I = 2\pi \sqrt{2} - 2\pi \implies I = \pi(\sqrt 2 - 1)$$ — Brevan Ellefsen, May 12 '19 at 08:50
Possible duplicate of computing $\int _{-1}^1:\frac{\sqrt{1-x^2}}{x^2+1}$ using residue calculus — Martin Hansen, May 14 '19 at 16:40

Integrate $\frac{\sqrt{1-x^2}}{1+x^2}$ from $-1$ to $1$ using the residue theorem

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