Answer
$\tan^{-1} |x|+C$
Work Step by Step
Consider $x= \tan \theta \implies dx=\sec^2 \theta d\theta$
Now, the given integral can be written as:
$\int\dfrac{\sec ^2 \theta}{1+\tan^2 \theta } d\theta= \int \dfrac{\sec^2 \theta}{\sec^2 \theta} d\theta$
or, $= \int d \theta$
or, $=\tan^{-1} |x|+C$
