Find

t$2$e$9$t dt$.$

Solution

Notice that t$2$ becomes simpler when differentiated $($whereas

e$9$t

is mostly unchanged when differentiated or integrated$),$ so we choose

u $=$ t$2$ and dv $=$ e$9$t dt$.$

Then

du $=$

dt

and

v $=$

$.$

Integration by parts gives

t$2$e$9$t dt $=$

$$

$29$

te$9$t dt $(3).$

The integral that we obtained,

te$9$t dt$,$

is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time, this time with

u $=$ t

and

dv $=$ e$9$t dt$.$

Then

du $=$ dt$,$

v $=$

$,$

and

te$9$t dt $=$

$19$

$$te$9$t $$

$$dt $=$

$+$ C$.$

Putting this in Equation $(3),$ we get

t$2$e$9$t dt

$=$

$19$

$$t$2$e$9$t $$

te$9$t dt

$=$

$19$

$$t$2$e$9$t $$

$$

$19$

te$9$t $$

$181$

e$9$t $+$ C

$=$

$+$ C$1,$

where

C$1=$

$29$

C