Find $f(1)+\cdots+f(60)$ 
Let $f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$. Find $f(1)+\cdots+f(60)$.

I considered rationalizing the denominator, but that seems to make the fraction more complicated. We get $$\dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} = \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right).$$ is there an easier way?

Maybe it's easier if you think that
$$\begin{align}
A = \sqrt{2x+1}&\qquad B = \sqrt{2x-1}\\
AB &= \sqrt{4x^2-1}\\
A^2 + B^2 &= 4x
\end{align}$$
When you split them you get
$$A+B -\frac{AB}{A+B}$$
Then conjugate the right term with $A-B$
$$A+B - \frac{AB(A-B)}2\\
A+B-\frac{A^2B}2+\frac A{2B^2}$$
Now it looks more nicely seperated.
