{4 - 5e^ - 5t + (e^2t)/3}dt
I got
4t + e^-5t + (e^2t)/6
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