I am kinda of looking for some step by step of this problem below because I am having some trouble grasping parts.
$\displaystyle \int t^2 e^(3t) dt$
Thanks, Tyler
PS: 3t is all in e^, I guess that didn't show up well.
Hello, Tyler!
If you're "having some trouble grasping parts",
. . this may be over your head . . .
$\displaystyle I \;=\;\int t^2 e^{3t} \,dt$
By parts: .$\displaystyle \begin{array}{ccccccc}
u &=& t^2 && dv &=& e^t\,dt \\
du &=& 2t\,dt && v &=& e^t \end{array}$
Then: .$\displaystyle I \;=\;t^2e^t - 2\!\!\int te^t\,dt$
By parts again: .$\displaystyle \begin{array}{ccccccc}u &=& t && dv &=& e^t\,dt \\
du &=& dt && v &=& e^t \end{array}$
Then: .$\displaystyle I \;=\;t^2e^t - 2\bigg[te^t - \int e^t\,dt\bigg] \;=\;t^2e^t - 2\bigg[te^t - e^t + C\bigg] \;=\;t^2e^t - 2te^t + 2e^t + C $
Therefore: .$\displaystyle I \;=\;e^t(t^2 - 2t + 2) + C$