Integration by parts

**Latszer** · January 19th 2010, 08:36 PM

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.

**Soroban** · January 19th 2010, 08:56 PM

Hello, Tyler!

If you're "having some trouble grasping parts",
. . this may be over your head . . .

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

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

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

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

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

**Latszer** · January 19th 2010, 09:00 PM

why would v be just e^t and not e^3t?

**Jhevon** · January 19th 2010, 09:13 PM

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.

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