# Integrate?

#### Warrenx

(t^2 + e^(2t) + e^(-2t))^1/2

anyone got any ideas, I suck at integration so its hard for me to see what to do, if someone can start me in the right track I am sure I can finish.

thanks,

Warren.

#### p0oint

Hello!

Is the original integral that you want to solve:

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

#### Warrenx

Hello!

Is the original integral that you want to solve:

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

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

#### Warrenx

Dear Warrenx,

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

MHF Hall of Honor
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 .

• Warrenx

#### Warrenx

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... but thank you!