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?
 A: $$f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$$ $$= \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)$$
Now say $a=\sqrt{2x+1}$ and $b=\sqrt{2x-1}$
Then $f(x)=\dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right)=\frac{1}{2}\left(a^2+b^2+ab\right)\left(a-b\right)=\frac{1}{2}(a^3-b^3)$
Therefore, $$\boxed{f(x)=\frac{1}{2}\left[(2x+1)^\frac{3}{2}-(2x-1)^\frac{3}{2}\right]}$$
So, $f(1)+\cdots+f(60)=\frac{1}{2}\left[(2\cdot 60+1)^\frac{3}{2}-(2\cdot 1-1)^\frac{3}{2}\right]=665$
Hope this helps you.
A: Using the fact that $(\sqrt{2x+1}+\sqrt{2x-1})^2=4x+2\sqrt{4x^2-1}$, you can produce :
\begin{align}
  \frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} &=
  \frac{(\sqrt{2x+1}+\sqrt{2x-1})^2-\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}(\sqrt{2x+1}-\sqrt{2x-1})}{2} \\ 
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{(2x+1)}{2}\sqrt{2x-1} + \frac{(2x-1)}{2}\sqrt{2x+1} \\ 
&= \frac{(2x+1)^{3/2}}{2} - \frac{(2x-1)^{3/2}}{2}
\end{align}
So your sum is telescopic, and you find :
$$\sum_{k=1}^{60} f(k) = \frac{(2\times 60+1)^{3/2}}{2} - \frac{(2\times 1-1)^{3/2}}{2} = 665$$
Note : I'm sure there's a much cleaner way to obtain this, but I don't see how :-)
A: 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.
