$\displaystyle \int_{0}^{\infty} \dfrac{dx}{x^{2} + 1}$

$\displaystyle x = (1)\tan\theta$

$\displaystyle dx = \sec^{2}\theta d\theta$

$\displaystyle (\tan\theta)^{2} + 1$

$\displaystyle \tan^{2}\theta + 1 = sec^{2}\theta$

$\displaystyle \int_{0}^{\infty} \dfrac{\sec^{2}\theta}{\sec^{2}\theta}$ - ??? Simplify further or take integral? Simplification seems to just make a 1 Maybe trig substitution wasn't the way to go.