Yes that is it. I am sorry I did not use Latex, have headache right now and couldn't thinkHello!

Is the original integral that you want to solve:

\(\displaystyle \int \sqrt{t^2 + e^{2t} + e^{-2t}} dt\)

Please use LaTeX.

Dear Warrenx,Yes that is it. I am sorry I did not use Latex, have headache right now and couldn't think

This integration could not be evaluated using elementary functions. You can get the answer using, http://wims.unice.fr/wims/wims.cgi?session=4R29E4903D.2&+lang=en&+module=tool/analysis/function.en

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hmm... I think I wrote the problem write:Dear Warrenx,

This integration could not be evaluated using elementary functions. You can get the answer using, Function calculator

It is asking for the arc length:

so s(t) =\(\displaystyle \int_0^1 |r'(t)|\)

r(t) = \(\displaystyle \sqrt2t , e^t , e^{-t}\)

so I get r'(t) = \(\displaystyle t , e^t , e^{-t}\)

then lr'(t)l = \(\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}\)

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck

hmm... I think I wrote the problem write:

It is asking for the arc length:

so s(t) =\(\displaystyle \int_0^1 |r'(t)|\)

r(t) = \(\displaystyle \sqrt2t , e^t , e^{-t}\)

so I get r'(t) = \(\displaystyle t , e^t , e^{-t}\)

then lr'(t)l = \(\displaystyle \sqrt{(t)^2 + (e^t)^2 + (-e^{-t})^2}\)

(sorry it was -e for the third term)

and then I am supposed to integrate it, but I suck at integratin' so I are stuck

Oh , the derivative of \(\displaystyle \sqrt{2} t\) is \(\displaystyle \sqrt{2} \)

so what the integral you are going to evaluate should be :

\(\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt \)

Trick is also found

\(\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2 \) (Happy)

the heaven is just in front of you now .

If you don't tell anyone about this, I wont tell anyone...Oh , the derivative of \(\displaystyle \sqrt{2} t\) is \(\displaystyle \sqrt{2} \)

so what the integral you are going to evaluate should be :

\(\displaystyle \int \sqrt{ \sqrt{2}^2 + e^{2t} + e^{-2t} }~dt \)

Trick is also found

\(\displaystyle 2 + e^{2t} + e^{-2t} = (e^t)^2 + 2 (e^t)(e^{-t}) + (e^{-t})^2 = (e^t + e^{-t} )^2 \) (Happy)

the heaven is just in front of you now .

but thank you!

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