Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$? 
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$

I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?

It's a bit of overkill, but since $f(x) = \frac{1}{1+x^{2}}$ is a monotonically decreasing function on $(0, \infty)$, the Riemann-like sum $$\frac{1}{n}\sum_{k=0}^{{\color{red}{\infty}}} f \left(\frac{k}{n} \right)= \frac{1}{n} \sum_{k=0}^{\infty} \frac{1}{1+ \left(\frac{k}{n} \right)^{2}} = n\sum_{k=0}^{\infty} \frac{1}{n^{2}+k^{2}} $$ converges to the value of $\int_{0}^{\infty} \frac{dx}{1+x^{2}}$ as $n \to \infty$.
And the partial fraction expansion of  $\coth (\pi z)$ is  $$\coth(\pi z) = \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=1}^{\infty} \frac{1}{z^{2}+k^{2}} = - \frac{1}{\pi z} + \frac{2z}{\pi} \sum_{k=0}^{\infty} \frac{1}{z^{2}+k^{2}} .   $$
Therefore, $$ \begin{align}\int_{-\infty}^{\infty} \frac{dx}{1+x^{2}} &= 2 \int_{0}^{\infty} \frac{dx}{1+x^{2}} \\ &= 2 \lim_{n \to \infty} \left(\frac{1}{2n} + \frac{\pi}{2} \, \coth(\pi n) \right) \\ &= 2 \left(0+ \frac{\pi}{2}(1) \right) \\ &= \pi. \end{align}$$
