Answer
$\dfrac{1}{2}\ln (x^2+\sqrt{x^4+1})+C$
Work Step by Step
Consider $x^2=\tan \theta \implies 2x dx= \sec^2 \theta d \theta $
Now, the given integral can be written as:
$\dfrac{1}{2}\int\dfrac{\sec^2 \theta}{ \sqrt{1 +\tan^2 \theta}} d\theta= \dfrac{1}{2} \int \sec \theta d\theta$
or, $=\dfrac{1}{2}\ln |\sec d \theta+\tan d \theta|+C$
or, $=\dfrac{1}{2}\ln (x^2+\sqrt{x^4+1})+C$
