# Find integral of 3x^2 e^2x^3 for x

1. Jul 5, 2008

### staples82

1. The problem statement, all variables and given/known data
Use substitution to find: (3x^2)*(e^2x^3)*(dx)

I don't know how to do the integration sign...
2. Relevant equations

3. The attempt at a solution
u=2x^3
du=6x^2

this is where I got confused, my teacher isn't that good and I'm trying to understand how to substitute...

I know that its supposed to be e^u*du=e^u+c

2. Jul 5, 2008

### rock.freak667

Re: Integration

$$\int 3x^2 e^{2x^3}dx$$

$u=2x^3 \rightarrow \frac{du}{dx}=6x^2 \rightarrow du=6x^2 dx$

which is the same as $\frac{du}{2}=3x^2 dx$

You understand up to here right?

Now you can write the integral as this

$$\int e^{2x^3} (3x^2 dx)$$

Now do you see what you can put at $2x^3$ and $(3x^2 dx)$ as?

3. Jul 5, 2008

### staples82

Re: Integration

I'm wondering why you did du/2=3x^2dx, also how did you get the math text?

4. Jul 5, 2008

### rock.freak667

Re: Integration

Well you see, when you did the substitution of $u=3x^2$, a 'u' will appear to be integrated and you can't integrate 'u' with respect to x. So what we need to get something to replace 'dx' with and $\frac{du}{dx}=6x^2$, if we multiply by the 'dx' we'll see that $du=6x^2 dx$. But in the integrand there is no '$6x^2$', only $3x^2$. BUT $6x^2=2(3x^2)$, so that is why we divided by 2.

just use [*tex]What you want to represent mathematically[/tex*] without the *

5. Jul 5, 2008

### staples82

Re: Integration

Ok, so then thats why $$e^u/2$$ is the answer, I think I have to look at a few more problems, see if i can clear it up

6. Jul 5, 2008

### rock.freak667

Re: Integration

Post back if you need more help.