Difficult integral: $\int_0^{\ln2} {\sqrt{e^{2x} -2 +e^{-2x}}\over e^{x}+e^{-x}}\,dx$ $$\int_0^{\ln2} {\sqrt{e^{2x} -2 +e^{-2x}}\over e^{x}+e^{-x}}\,dx$$
This integral is very difficult, and I don't know how to solve it or even start with it. I'm sorry for not being able to write this in a better way; if someone can do that for me it would also be great. But I hope you can understand what I basically want help with.
 A: HINT: Rewrite $$(e^x-e^{-x})^2.$$
A: Hint: Although there are better suggestions, note that $$e^x+e^{-x}=2\cosh x$$ where the function $\cosh$ has the following properties (that you will need, if you follow this way, which you do not need to, since there are faster ways suggested):


*

*$\cosh 2x=\cosh^2x+\sinh^2x$, 

*$\cosh^2 x-\sinh^2x=1$,

*$(\cosh x)'=\sinh x$ and $(\sinh x)'=\cosh x$

A: Note that $e^{2x}-2+e^{-2x}=(e^x-e^{-x})^2$. 
Then you will find $u=e^x+e^{-x}$ useful.
Added: We want to find
$$\int_{x=0}^{\ln 2} \frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx.$$
Let $u=e^x+e^{-x}$. Note that $du=(e^x-e^{-x})\,dx$, and conveniently $(e^x-e^{-x})\,dx$ happens to be sitting on top. Substituting, we get
$$\int_{u=2}^{\frac{5}{2}}\frac{1}{u}\,du.$$
Integrate. We get $\ln(5/2)-\ln(2)$, which can also be written as $\ln(5/4)$. 
Remark: Alternately, we could express the general antiderivative in terms of $x$, and then do the calculation. The general antiderivative  is $\ln(e^x+e^{-x})+C$.  
A: $$ \sqrt{(e^x-e^{-x})^2}=|e^x-e^{-x}|$$
but you first you need determinate. If $$ e^x-e^{-x}$$ is positive or negative because if this term was negative then the absolute value would equal to $ -( e^x-e^{-x}) $
but in this case it's positive, then the algebraic expresion stays intact
best regards
