Answer
$\ln|\dfrac{5x}{3} +\dfrac{\sqrt {25x^2-9}}{3}|+C$
Work Step by Step
Let us consider $x =\dfrac{3}{5} \sec \theta $
and $dx=\dfrac{3}{5} \sec\theta \tan \theta d \theta $
Then
$\int \dfrac{5(\dfrac{3}{5} \sec\theta \tan \theta d \theta)}{\sqrt{25(\dfrac{9}{25} \sec^2 \theta-9)}} = \int \sec \theta d \theta=\ln |\sec\theta + \tan \theta|+C$
Substitute $\sec \theta =\dfrac{5x}{3}$
$ \ln |\sec\theta + \tan \theta|+C= \ln|\dfrac{5x}{3} +\dfrac{\sqrt {25x^2-9}}{3}|+C$
