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