Maybe a trivial problem Can anyone give me some suggestion about an elementary method (that is, without using advanced theorems in complex analysis. I only know that to integrate a complex-valued function, we do it with respect to its real part and imaginary part respectively. I did not learn complex analysis before...) to show that the following integration is equal to $0$? Many thanks!
$$\int_\mathbb{R} \frac{1}{1+x^2}\cdot \bigg(\frac{i-x}{i+x}\bigg)^n\,\mathrm{d} x$$
where $n\in \mathbb{N}$. 
 A: The crazy thing is this (which I got after seeing WA results):
\begin{align*}
\frac d {dx}\left(\frac{i-x}{i+x}\right)^n&=n\left(\frac{i-x}{i+x}\right)^{n-1}\frac{-(i+x)-(i-x)}{(i+x)^2}\\
&=n\left(\frac{i-x}{i+x}\right)^{n-1}\frac{-2i}{(i+x)^2}\\
&=\left(\frac{i-x}{i+x}\right)^n\frac{-2in}{(i+x)(i-x)}\\
&=\left(\frac{i-x}{i+x}\right)^n\frac{2in}{1+x^2}
\end{align*}
So
$$\int \left(\frac{i-x}{i+x}\right)^n\frac{2in}{1+x^2}\,dx=\left(\frac{i-x}{i+x}\right)^n+C_0,$$
so
$$\int \frac 1 {x^2+1}\left(\frac{i-x}{i+x}\right)^n\,dx=\frac1{2in}\left(\frac{i-x}{i+x}\right)^n+C.$$
\begin{align*}
\frac{i-x}{i+x}
&=\frac{i^2-2ix+x^2}{i^2-x^2}\\
&=\frac{x^2-1-2ix}{-1-x^2}.
\end{align*}
As $x\to\pm\infty$, the real part of this ratio approaches $-1$ while the imaginary part approaches $0$. This same behavior holds, then, for all positive integer powers of the ratio, so the definite integral over the reals is $0$ as you claimed.
Note: a somewhat more interesting sort of odd-even pattern appears if you enter the integral incorrectly, taking it from $0$ to $\infty$ instead of $-\infty$ to $\infty$.
