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

[misogynisticfeminist]

Hi, I've actually got a problem here.

How do I evaluate

\int e^x^3 x^2 dx

I have problem when doing integration by parts of finding \int v du 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

\int e^x^3 x^2 dx

I have problem when doing integration by parts of finding \int v du 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:
\int e^{x^3} x^2 dx.
If so,this integral is trivial and it does not require anything,not even the lousy substitution x^3 =u.
So:
\int e^{x^3} x^2 dx =\frac{1}{3} e^{x^3} +C.
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 u = x^2 and dv = e^{3x} dx

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

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

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

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

\frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{2}{27}e^{3x}


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


Edit: Added LateX. That takes forever.