Is there a way to do this from the second expression? I ended up where I started from. I do not want to use trigonometrical substitution.
There is one way to solve it without trig substitutiton but it does not involve the second expression.
$\displaystyle \int_{5/4}^{5/3}\frac{du}{\sqrt{u^2-1}}$
$\displaystyle =\int_{5/4}^{5/3}\frac{(u+\sqrt{u^2-1})}{(u+\sqrt{u^2-1})\sqrt{u^2-1}}du$
$\displaystyle =\int_{5/4}^{5/3}\frac{\frac{u}{\sqrt{u^2-1}}+1}{(u+\sqrt{u^2-1})}du$
(This is $\displaystyle \frac{f'(u)}{f(u)}$ form)
$\displaystyle =[\ln{(u+\sqrt{u^2-1})}]_{5/4}^{5/3}$
Note- The identity could also have been used :$\displaystyle \int \frac{1}{\sqrt{x^2-a^2}}dx=\ln{(x+\sqrt{x^2-a^2})}+C$
Look, stuck man, I donot think there is any way to find the integral from the 2 fractions. Even wolfram alpha solves it using trigonometric substitution. You can see it here if you like. link. If required you can also use hyperbolic substitution.