# Integration by parts

• January 19th 2010, 09:36 PM
Latszer
Integration by parts
I am kinda of looking for some step by step of this problem below because I am having some trouble grasping parts.

$\int t^2 e^(3t) dt$

Thanks, Tyler

PS: 3t is all in e^, I guess that didn't show up well.
• January 19th 2010, 09:56 PM
Soroban
Hello, Tyler!

If you're "having some trouble grasping parts",

Quote:

$I \;=\;\int t^2 e^{3t} \,dt$

By parts: . $\begin{array}{ccccccc}
u &=& t^2 && dv &=& e^t\,dt \\
du &=& 2t\,dt && v &=& e^t \end{array}$

Then: . $I \;=\;t^2e^t - 2\!\!\int te^t\,dt$

By parts again: . $\begin{array}{ccccccc}u &=& t && dv &=& e^t\,dt \\
du &=& dt && v &=& e^t \end{array}$

Then: . $I \;=\;t^2e^t - 2\bigg[te^t - \int e^t\,dt\bigg] \;=\;t^2e^t - 2\bigg[te^t - e^t + C\bigg] \;=\;t^2e^t - 2te^t + 2e^t + C$

Therefore: . $I \;=\;e^t(t^2 - 2t + 2) + C$

• January 19th 2010, 10:00 PM
Latszer
why would v be just e^t and not e^3t?
• January 19th 2010, 10:13 PM
Jhevon
Quote:

Originally Posted by Latszer
why would v be just e^t and not e^3t?

a typo that was carried through the whole solution. it shouldn't be difficult for you to go through and put in the 3's and 1/3's everywhere they should go.