Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 3

$\dfrac{\pi}{4}$

Consider $x =2\tan \theta \implies dx= 2 \sec^2 \theta d \theta $ Then, $\int \dfrac{2 \sec^2 \theta d \theta }{\sqrt {4+4 \tan^2 \theta)}} = \int (\dfrac{1}{2}) d \theta$ Now, $(\dfrac{1}{2}) \int_{-2}^{2} d \theta=\dfrac{1}{2}[\tan^{-1} (1)-\tan^{-1} (-1)]$ Thus, $\int \dfrac{2 \sec^2 \theta d \theta }{\sqrt {4+4 \tan^2 \theta)}} =\dfrac{\pi}{4}$