{4 - 5e^ - 5t + (e^2t)/3}dt

I got

4t + e^-5t + (e^2t)/6

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- Jul 1st 2008, 09:58 PM10164064is this correct?
{4 - 5e^ - 5t + (e^2t)/3}dt

I got

4t + e^-5t + (e^2t)/6 - Jul 1st 2008, 10:06 PMMathstud28
- Jul 1st 2008, 11:00 PMCaptainBlack
You can check this yourself by differentiating $\displaystyle 4t + e^{-5t} + e^{2t}/6$, you should get your integrand $\displaystyle 4 - 5e^{- 5t} + e^{2t}/3$.

Also since this is an indefinite integralb there should be an arbitary constant added so you should have:

$\displaystyle 4t + e^{-5t} + e^{2t}/6+c$

RonL