Question

indefinite integral of x^3e^x^2dx ?

Answer 1

\[\int\limits x^3e^{x^2}dx\] ?
I know it's integration by parts...

Answer 2

would you do u = e^x^2 and dv = x^3 ?

Answer 3

try letting u=x^3 and dv = e^(x^2)

Answer 4

sorry other way around lol
dv = x^3 and u = e^(x^2)

Answer 5

i did that before...
so would you have to do integration by parts like two or three times?

Answer 6

so v = 1/4 x^4
and du = 2xe^(x^2) dx

Answer 7

then using the equation uv - \[\int\limits_{}^{} \] vdu
you can use a U-substitution for that integral

Answer 8

lol bad formatting its uv - integral of vdu

Answer 9

okay, so putting all that in, i get
\[\int\limits x^3 e^{x^2} dx = \frac{ e^{x^2}x^4 }{ 4 } - 2 \int\limits x^4e^{x^2}dx\]

Answer 10

how would you go from there?

Answer 11

sorry mistake do a u-sub first so let u =x^2

Answer 12

i dont know if this is correct but.

integral (x^3)(e^x^2)dx
u = x^2
du = 2xdx
du/2x = dx

integral x(x^2)(e^u)(du/2x)

so you would get
intl u(e^u)du
so...that would proll make it a little better. for IBP

Answer 13

once you do the u-sub it will become \[\int\limits_{}^{} ue^u du \]

Answer 14

and just use integration by parts from theere

Answer 15

i did the u = x^2
and i got
\[ 2 \int\limits x^4e^{x^2}dx = 2(\frac{ x^7 }{ 3 } - \frac{ 4x^5 }{ 15 }) +C\]

Answer 16

so u = x^2, du=2xdx so xdx=du/2
\[\int\limits_{}^{} x^2 e^u du\]
and since u =x^2 you replace that x^2 with a u

Answer 17

my bad theres also a 1/2 in that integral

Answer 18

so it just becomes \[\frac{ 1 }{ 2 } \int\limits_{}^{}ue^u du\]
and you use integration by parts from here

Answer 19

int (1/2 u e^u) du
use the tabular integration by part