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.

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- Jan 19th 2010, 08:36 PMLatszerIntegration by parts
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. - Jan 19th 2010, 08:56 PMSoroban
Hello, Tyler!

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

. . this may be over your head . . .

Quote:

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

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

- Jan 19th 2010, 09:00 PMLatszer
why would v be just e^t and not e^3t?

- Jan 19th 2010, 09:13 PMJhevon