How do I integrate e^(x^3) x^2 dx?

[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.