## University Calculus: Early Transcendentals (3rd Edition)

$\tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$
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: $(\theta+\sin \theta \cos \theta) +C= \tan^{-1} (3t) +\dfrac{3t}{(9t^2+1)}+C$