# Thread: is this correct?

1. ## is this correct?

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

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

2. Originally Posted by 10164064
{4 - 5e^ - 5t + (e^2t)/3}dt

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

3. Originally Posted by 10164064
{4 - 5e^ - 5t + (e^2t)/3}dt

I got
4t + e^-5t + (e^2t)/6
You can check this yourself by differentiating $4t + e^{-5t} + e^{2t}/6$, you should get your integrand $4 - 5e^{- 5t} + e^{2t}/3$.

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

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

RonL