## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{\pi}{2}$
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, $\lim\limits_{a \to \infty} [\tan^{-1} a-\tan^{-1} 0]= \dfrac{\pi}{2}$