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?

You can use partial fractions:
$$
\begin{align}
\int_{-\infty}^\infty \frac{dx}{1+x^2}
& = \int_{-\infty}^\infty \frac{1}{2} \left( \frac{1}{1+ix} + \frac{1}{1-ix} \right) dx \\
& = \frac{1}{2i} \bigg[\log(1+ix) - \log(1-ix)\bigg]_{-\infty}^\infty \\
& = \frac{1}{2i} \left[ \lim_{x\to\infty} \log\left( \frac{1+ix}{1-ix} \right) 
 - \lim_{x\to-\infty} \log\left(\frac{1+ix}{1-ix}\right)\right] \\
& = \frac{1}{2i}\bigg[ i\pi - (-i\pi)\bigg] = \pi
\end{align}
$$
Note that the expression on the second line obviously must be equal to
$$
\bigg[ \arctan x\bigg]_{-\infty}^\infty
$$
because we know the antiderivative of $1/(1+x^2)$ is $\arctan x$.
