Integral of $\displaystyle

(e^x-e^-x)/(e^x+e^-x)

$

I have tried adding the conjugate to the top and bottom but don't know where to go from there. Should I continue that or try a different technique?

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- Dec 2nd 2009, 06:32 PMstevo970integral
Integral of $\displaystyle

(e^x-e^-x)/(e^x+e^-x)

$

I have tried adding the conjugate to the top and bottom but don't know where to go from there. Should I continue that or try a different technique? - Dec 2nd 2009, 06:54 PMSoroban
Hello, stevo970!

If you look at it the right way, it's easy . . .

Quote:

$\displaystyle \int\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx$

Let $\displaystyle u$ = the denominator.

We have: .$\displaystyle \begin{array}{ccc}u &=& e^x + e^{-x} \\ du &=& (e^x - e^{-x})\,dx \end{array}$

Hence, we have: .$\displaystyle \int\frac{du}{u}$

Got it?

- Dec 3rd 2009, 11:42 AMstevo970
wow ok thank you

- Dec 3rd 2009, 12:06 PMhmmmm
also noting that this is (sinh(x))/(cosh(x)) may make this become clearer