Answer
$(\sqrt {y^2-49}-7 \sec^{-1} (y/7))+C$
Work Step by Step
Let us consider $y = \sec \theta $
and $dy=7\sec\theta \tan \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{\sqrt{(49\sec^2 \theta-49)}}{7 \sec \theta} = 7 \int (\sec^2 \theta-1) d \theta$
Now, integrate and plug in $\sec \theta =(y/7)$:
$ 7( \tan \theta -\theta)+C=(\sqrt {y^2-49}-7 \sec^{-1} (y/7))+C$