Length of the curve integral, can't solve the integral

[Prometheos]

This question was on my test I have no idea how to do the middle work.

Find the length of the curve
y = \frac{1}{2}(e^x + e^{-x}) , 0 \leq x \leq 2

Problem set up was easy enough
L= \int_0^2 \sqrt{ 1 + \frac{1}{4}( e^{2x} -2 + e^{-2x} ) } dx

Looking back in my notes I see that the answer is
\frac{1}{2}( e^2 - e^{-2} )

But, how do you get there? I think my main problem is probably the algebra behind combing the 1 and derivative of y squared.

 

[Dick]

Combine the 1 and (1/4)(-2) constants under the radial. You can now rearrange into a perfect square. Remember e^(2x)=(e^x)^2.

 

[Prometheos]

Ah, so you get \frac{1}{4}e^<br /> {2x} + \frac{1}{2} + \frac{1}{4}e^{-2x}
after distributing the .25 and adding the one which is equal to
\frac{1}{4} ( e^{x} + e^{-x} )^2
Wow, high school algebra comes back to haunt me again.

Thanks for the help, I may have missed this problem on the test, but hopefully it won't happen again now lol