Stuck on Integrating e^(x^3) x^2?

[misogynisticfeminist]

Hi, I've actually got a problem here.

How do I evaluate

[tex]\int e^x^3 x^2 dx[/tex]

I have problem when doing integration by parts of finding [tex] \int v du [/tex] since if I integrate v du, i'll get another expression which i have to integrate by parts again, and this goes on and on !


(its meant to be e to the power x cubed by the way).

 

[dextercioby]

Hi, I've actually got a problem here.

How do I evaluate

[tex]\int e^x^3 x^2 dx[/tex]

I have problem when doing integration by parts of finding [tex] \int v du [/tex] since if I integrate v du, i'll get another expression which i have to integrate by parts again, and this goes on and on !


(its meant to be e to the power x cubed by the way).

It probably reads:
[tex]\int e^{x^3} x^2 dx[/tex].
If so,this integral is trivial and it does not require anything,not even the lousy substitution [tex] x^3 =u [/tex].
So:
[tex]\int e^{x^3} x^2 dx =\frac{1}{3} e^{x^3} +C [/tex].
If it's not as i interpreted it,well,then 2 integrals by parts should simply do the trick.
Good luck!

 

[WaR]

Hello
It doesn't go on forever. You only do it twice.
The first time [tex]u = x^2[/tex] and [tex]dv = e^{3x} dx[/tex]

Then you get the following:
[tex] \frac{1}{3} x^2 e^{3x} - \int \frac{1}{3} e^{3x} 2x dx[/tex]

Now, you do integration by parts a second time
This time [tex]u = 2x[/tex] and [tex]dv = e^{3x} dx[/tex] (remember you can pull out that 1/3)

Then you get the following:
[tex] \frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \int \frac{1}{3} e^{3x} 2 dx[/tex]

The last integral doesn't need integration by parts, it's just a simple integration. You should get the following:

[tex]\frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27}e^{3x}[/tex]


Sorry for not using LateX. I'll get the hang of it later though :)


Edit: Added LateX. That takes forever.