QUAM Review Question 4 - Editing titles is fun

Let $\gamma$ be a simple, closed loop in the top half of the plane enclosing the point $1+i$. I am asked to compute
$$
\int_\gamma\frac{1}{z^4+1} \ dz.
$$
Using the Cauchy integral formula (evaluating at $z_0=1+i$), I get that this integral evaluates to $\pi/8+(\pi/8)i$.

Now, supposing that $\gamma$ is a semicircular arc, we are asked to compute
$$
\int_{-\infty}^\infty\frac{1}{x^4+1} \ dx.
$$
I was unable to do the integral, but Mathematica tells me it equals $\pi/\sqrt{2}$.

Are these the correct answers? And if not, can anybody tell me why?

Okay, so here's an update.

I re-evaluated the integral
$$
\int_\gamma\frac{1}{z^4+1} \ dz
$$
at the pole $z_0=1+i$ and got $\pi/8-(\pi/8)i$.

I then let $\gamma$ be a semi-circular arc and evaluated the integral at the poles $1+i$ and $-1+i$.

The gives $\pi/8-(\pi/8)i+\pi/8+(\pi/8)i=\pi/4$.

Even though this integral should correspond to
$$
\int_{-\infty}^\infty\frac{1}{x^4+1} \ dx,
$$
it still doesn't give the $\pi/\sqrt{2}$ that I was hoping for.

Any suggestions?

I think the issue lies in that this function doesn't have a pole at $1+i$ or $-1+i$; of its four poles, the two in the upper half plane are $(1+i)/\sqrt{2}$ and $(-1+i)/\sqrt{2}$. I think if you evaluate at these, you will have the right result.

Note: Do not trust factorizations or other hints given on review sheets.

Or exams.

And don't forget why math is better than history.