Answer
$\ln|\dfrac{5x}{3} +\dfrac{\sqrt {25x^2-9}}{3}|+C$
Work Step by Step
Let us consider $t =\dfrac{3}{5} \sec \theta $
and $dt=\dfrac{3}{5} \sec\theta \tan \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{5(\dfrac{3}{5} \sec\theta \tan \theta d \theta)}{\sqrt{25(\dfrac{9}{25} \sec^2 \theta-9)}} = \int \sec \theta d \theta$
Now, integrate and plug in: $\sec \theta =(5x/3)$
$ \ln |\sec\theta + \tan \theta|+C= \ln|\dfrac{5x}{3} +\dfrac{\sqrt {25x^2-9}}{3}|+C$