1. ## integration by parts

Use integration by parts to find the integral.

$\int t^2*\exp{3t} dt$
$u=t^2$ $v'=\exp{3t}$
$u'=2t$ $v=\frac{\exp{3t}}{3}$
$\frac{1}{3}t^2\exp{3t}-\int\frac{2t\exp{3t}}{3}dt =\frac{1}{3}t^2\exp{3t}-\frac{2}{3}\int t\exp{3t} dt$

$u=t$ $v'=\exp{3t}$
$u'=1$ $v=\frac{\exp{3t}}{3}$

$\frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\int\frac{1}{3}\exp{3t} dt)$
$\frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\frac{1}{3}*\frac{1}{3}\exp{3t})+C$

To clean this up a bit...
$\frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\frac{1}{3}*\frac{1}{3}\exp{3t})+C$

$\frac{1}{3}t^2\exp{3t}-\frac{2}{9}t\exp{3t}-\frac{1}{3}\exp{3t}+C=$

$\frac{\exp{3t}}{3}(t^2-\frac{2}{9}t-1)$

Did I do all of that alright? It was not very easy to check with differentiation.

2. The last term should be 2/29 e^(3t)

not that hard to differentiate