Answer
$${\tan ^{ - 1}}x + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 + {x^2}}}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}x = \tan \theta,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{dx}}{{1 + {x^2}}}} = \int {\frac{{{{\sec }^2}\theta }}{{1 + \tan {\theta ^2}}}d\theta } \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{1 + \tan {\theta ^2}}}\left( {{{\sec }^2}\theta } \right)d\theta } \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta } \cr
& {\text{simplifying}} \cr
& = \int {\theta d\theta } \cr
& {\text{Integrate}} \cr
& = \theta + C \cr
& {\text{write in terms of }}x,{\text{ }}x = \tan \theta \,\,\,\, \Rightarrow \,\,\,\theta = {\tan ^{ - 1}}x \cr
& = {\tan ^{ - 1}}x + C \cr} $$
