Answer
$ \dfrac{\pi}{2}$
Work Step by Step
Consider $f(x)=\int_{0}^{\infty} \dfrac{1}{x^2+1} dx$
$\lim\limits_{a \to \infty} f(x)= \lim\limits_{a \to \infty}\int_{0}^{a} \dfrac{1}{x^2+1} dx=\lim\limits_{a \to \infty} [\tan^{-1} x]_{0}^{a}$
or, $\lim\limits_{a \to \infty} [\tan^{-1} a-\tan^{-1} 0]= \dfrac{\pi}{2}$
