Calculate $ \int_{\mathbb{R}} \frac{dx}{x^4+1}$ using the residues theorem. Calculate using the residues theorem this integral : $$ \int \limits_{-\infty}^{+\infty} \frac{\mathrm{d}x}{x^4+1}.
$$

First I calculated $\displaystyle \int_{C_r} \frac{\mathrm{d}z}{z^4+1} $, where $C_r$ is a half circle centered in the origin and it radius $r>1$ including the real axis, I found (by summing the residues) that it equals : $\displaystyle \frac{\pi}{\sqrt{2}}$, considering : $T_r$ as the arc of $C_r$ (no real axis) I got : $$ \frac{\pi}{\sqrt{2}} = \int\limits_{-r}^{r} \frac{\mathrm{d}z}{z^4+1} + \int_{T_r}\frac{\mathrm{d}z}{z^4+1} .$$
Now, what can we say about $$\lim_{r\to \infty} \int_{T_r}\frac{\mathrm{d}z}{z^4+1}$$ It seem that it tends toward zero but I cant prove it. if you think that it is easy, then just post hints.
Some googling lead me to this, how do we prove it ? (the easiest method).
Thanks in advance.
 A: We have that
$$\text{On}\;\;C_r\,,\;\;|z|=r\implies \left|\;\int\limits_{C_r}\frac{1}{z^4+1}dz\;\right|\le\frac{1}{r^4-1}\pi r\xrightarrow[r\to\infty]{}0$$
A: The points of $T_r$ are of the form $z(\theta) = r\operatorname{e}^{i\theta}$ where $0 \le \theta \le \pi$. We can substitute this into the integral. Forst notice that $\operatorname{d}\!z = ir\operatorname{e}^{i\theta}\, \operatorname{d}\!\theta$, giving:
$$\int_{T_r} \frac{1}{z^4+1} \, \operatorname{d}\!z = \int_0^{\pi} \frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \, \operatorname{d}\!\theta $$
We can estimate the size of this integral:
\begin{array}{ccc}
\left|\int_0^{\pi} \frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \, \operatorname{d}\!\theta\right| &\le& \int_0^{\pi} \left|\frac{ir\operatorname{e}^{i\theta}}{r^4\operatorname{e}^{4i\theta}+1} \right| \operatorname{d}\!\theta \\
&=& \int_0^{\pi}\frac{r}{\left|r^4\operatorname{e}^{4i\theta}+1\right|}  \operatorname{d}\!\theta \\
&\le& \int_0^{\pi}\frac{r}{\left|r^4\operatorname{e}^{4i\theta}\right|-\left|1\right|}  \operatorname{d}\!\theta \\
&=& \int_0^{\pi}\frac{r}{r^4-1}  \operatorname{d}\!\theta
\end{array}
In step three I applied a version of the triangle inequality which says that, for all $z,w \in \mathbb{C}$ we have $|z+w| \ge ||z|-|w||$. As $r \to \infty$ this last integral tends towards zero. Hence our original internal terns to zero as $r \to \infty$.
