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$$\int_0^{\ln2} {\sqrt{e^{2x} -2 +e^{-2x}}\over e^{x}+e^{-x}}\,dx$$

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.

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trinomial inside the radical is factorable as a perfect square. – imranfat Mar 16 '14 at 18:28
Try the substitution $u=e^x$. I'm sure things will look cleaner from there. – ireallydonknow Mar 16 '14 at 18:29
@mookid. I thought the denominator was outside the radical... – imranfat Mar 16 '14 at 18:30
@imranfat: this way? – mookid Mar 16 '14 at 18:31
This integral is very difficult - No. It's not. It's the exact opposite, actually. Equally “difficult” integrals would be $\quad\displaystyle\int e^{\ln x}dx\quad$ or $\quad\displaystyle\int x^{^{\tfrac x{\ln x}}}dx$. – Lucian Mar 16 '14 at 18:40

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 $\ln(e^x+e^{-x})+C$.

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Can i then use u-substitution or do i have to use integration by parts? – user135839 Mar 16 '14 at 18:40
Substitution does it. You will end up with $\ln(e^x+e^{-x})+C$. – André Nicolas Mar 16 '14 at 18:41
What i have a problem with is that if i should use u-substitution it wouldn't be the same in the numerator and denominator. u=e^x+e^-x will not give same substitution as for e^x-e^-x – user135839 Mar 16 '14 at 18:48
I will write it out. – André Nicolas Mar 16 '14 at 18:49
Wow, i never thought of it that way. That was actually brilliant and made 100% sense. Thank you very much! :) – user135839 Mar 16 '14 at 19:01

HINT: Rewrite $$(e^x-e^{-x})^2.$$

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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):

  1. $\cosh 2x=\cosh^2x+\sinh^2x$,
  2. $\cosh^2 x-\sinh^2x=1$,
  3. $(\cosh x)'=\sinh x$ and $(\sinh x)'=\cosh x$
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$$ \sqrt{(e^x-e^{-x})^2}=|e^x-e^{-x}|$$ 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

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