# I dont know the code for exponential

1. Jan 3, 2006

### frozen7

$$\int \ell^{x^2} x^2 dx$$

$$= 1/2 \int \ell^{x^2} x^2 d(x^2)$$
$$= 1/2 \int \ell^{t} . t dt$$ (Let x^2 = t )
$$= 1/2 [t.\ell^{t} - \int \ell^{t} dt]$$
$$= 1/2 [t.\ell^{t} - \ell^{t} + c ]$$
$$= 1/2 [x^2 \ell^{x^2} - \ell^{x^2} + c]$$

$$\ell$$ is actually exponential. I dont know the code for exponential. :)
What`s the problem with the above integration I have done?

Last edited: Jan 3, 2006
2. Jan 3, 2006

### d_leet

Well you set $$t= x^2$$ but you don't have $$d(x^2)$$ in the original integral, just integrate by parts you don't really have to make a substitution.

3. Jan 4, 2006

### VietDao29

The error lies in line #2, you are changing dx into d(x2), that requires an x, that'll left you with:
$$\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2)$$ not $$\frac{1}{2} \int x ^ 2 e ^ {x ^ 2} d(x ^ 2)$$. Or you can do a little bit more slowly by letting t = x2. Just try it, and see if you can get to:
$$\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2)$$.
From here, you can do the same and end up with:
$$\frac{1}{2} \int x e ^ {x ^ 2} d(x ^ 2) = \frac{1}{2} \int x d(e ^ {x ^ 2})$$.
From here, do you know what to do next?