Answer
$\dfrac{1}{2} \ln|\dfrac{2x}{7} +\dfrac{\sqrt {4x^2-49}}{7}|+C$
Work Step by Step
Let us consider $t =\dfrac{7}{2} \sec \theta $
and $dt=\dfrac{7}{2} \sec\theta \tan \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{\dfrac{7}{2} \sec\theta \tan \theta d \theta}{\sqrt{4(\dfrac{49}{4} \sec^2 \theta-49)}} = \dfrac{1}{2} \int \sec \theta d \theta$
Now, integrate and plug in $\sec \theta =(2x/7)$:
$ \dfrac{1}{2} \ln |\sec\theta + \tan \theta|+C=\dfrac{1}{2} \ln|\dfrac{2x}{7} +\dfrac{\sqrt {4x^2-49}}{7}|+C$
