Question

integral x^3e^(x^2)dx
integration by parts?

Answer 1

yes, thrice. have fun

Answer 2

would u= \[e ^{x ^{2}}\]

Answer 3

oh i read it wrong. i thought it was
\[\int x^3e^xdx\] sorry

Answer 4

so you can integrate by parts still, but i think you only need to do it once

\[u=x^2,du = xdx\]

Answer 5

down the e^x^2 maybe

Answer 6

so you get
\[\int u e^u du\] and now parts will give answer in one step, or rather you only have to use parts once

Answer 7

get
\[ue^u-\int e^u du=ue^u - e^u \] and you are done

Answer 8

well that is wrong i forgot the
\[\frac{1}{2}\] out front

Answer 9

\[u=x^2,du = 2xdx,\frac{1}{2}du=xdx\] is more like it.

Answer 10

\[\int x^3e^{x^2}dx=e^{x^2}*\frac{1}{4}x^4-\int 2xe^{x^2}*\frac{1}{4}x^4\]

\[\int x^3e^{x^2}dx=\frac{1}{4}x^4e^{x^2}-\frac{1}{2}\int x^3e^{x^2}dx\]
\[\frac{1}{2}\int x^3e^{x^2}dx+\int x^3e^{x^2}dx=\frac{1}{4}x^4e^{x^2}\]
\[\frac{3}{2}\int x^3e^{x^2}dx=\frac{1}{4}x^4e^{x^2}\]
\[\int x^3e^{x^2}dx=\frac{2}{3}\left(\frac{1}{4}x^4e^{x^2}\right)\]
\[\int x^3e^{x^2}dx=\frac{1}{6}x^4e^{x^2}\]
maybe?

Answer 11

messd that one up somewhere

Answer 12

are you making this one go around in a circle to get an equation? it might work!

Answer 13

yep

Answer 14

x*x^4 = x^5; not x&3 ..... thats my fauxpaux

Answer 15

but you are the master of the by parts method going down the diagonal as i recall, so it should be easier to use
\[u=x^2,\frac{1}{2}du=xdx\] and integrate
\[\frac{1}{2}\int u e^u du\] right?

Answer 16

those pesky laws of exponents...

Answer 17

the table works good; but i get to the circle thing here and get issues that I have to resolve thru therapy :)

Answer 18

:)

Answer 19

alright i think i got it

Answer 20

you should end up with
\[\frac{1}{2}(ue^u-e^u)\] and replace u by x^2 and that will be done

Answer 21

awsomeness. Thanks

Answer 22

i dont understand how you went from \[\int\limits {x^3}e ^{x ^{2}}dx = 1/6 x ^{4}e ^{x ^{2}} and then
1/2(ue ^{u}-e ^{u})\]