Answer
$\tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$
Work Step by Step
Let us consider $t = \dfrac{1}{3} \tan \theta $
and $dt= \dfrac{1}{3} \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{6( \dfrac{1}{3} sec^2 \theta)}{\sec ^4 \theta} d\theta=2\int cos^2 \theta d \theta $
Now, integrate:
$(\theta+\sin \theta \cos \theta) +C= \tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$
