Answer
$${\tan ^{ - 1}}3t + \frac{{3t}}{{9{t^2} + 1}} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{6dt}}{{{{\left( {9{t^2} + 1} \right)}^2}}}} \cr
& = 6\int {\frac{1}{{{{\left( {{{\left( {3t} \right)}^2} + 1} \right)}^2}}}} dt \cr
& {\text{Use the trigonometric substitutions}} \cr
& {\text{ }}3t = \tan \theta,{\text{ }}dt = \frac{1}{3}{\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& 6\int {\frac{{dt}}{{{{\left( {{{\left( {3t} \right)}^2} + 1} \right)}^2}}}} = 6\int {\frac{1}{{{{\left( {{{\tan }^2}\theta + 1} \right)}^2}}}\left( {\frac{1}{3}{{\sec }^2}\theta d\theta } \right)} \cr
& {\text{Use the fundamental identity }}\cr
& {\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& = 6\int {\frac{1}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}\left( {\frac{1}{3}{{\sec }^2}\theta d\theta } \right)} \cr
& {\text{simplifying, we get:}} \cr
& = 2\int {\frac{1}{{{{\sec }^4}\theta }}\left( {{{\sec }^2}\theta d\theta } \right)} \cr
& = 2\int {{{\cos }^2}\theta d\theta } \cr
& = 2\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& = \int {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& {\text{integrate}} \cr
& = \theta + \frac{1}{2}\sin 2\theta + C \cr
& = \theta + \sin \theta \cos \theta + C \cr
& {\text{Write in terms of }}x,\cr
& \sin \theta = \frac{{3t}}{{\sqrt {9{t^2} + 1} }},\,\,\,\cr
& \cos \theta = \frac{1}{{\sqrt {9{t^2} + 1} }}{\text{ and }}\cr
& \theta = {\tan ^{ - 1}}3t \cr
& = {\tan ^{ - 1}}3t + \left( {\frac{{3t}}{{\sqrt {9{t^2} + 1} }}} \right)\left( {\frac{1}{{\sqrt {9{t^2} + 1} }}} \right) + C \cr
& = {\tan ^{ - 1}}3t + \frac{{3t}}{{9{t^2} + 1}} + C \cr} $$
