Hi,

I don't know what I'm doing wrong with computing this integral. The book's answer is $\displaystyle \int (x^3)(e^{x^2}) dx = \frac{1}{2}e^{x^2}\left(x^2 - 1) + c $

I'll show my steps and then elaborate on what I think I've done wrong so far towards the end of my post.

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

$\displaystyle u = x^3 \Rightarrow \frac{du}{dx} = 3x^2 $

$\displaystyle \frac{d}{dx} (e^{x^2}) \Rightarrow v = \frac{1}{2x} e^{x^2} + c $

$\displaystyle \int (x^3)(e^{x^2}) dx = (x^3)\left(\frac{1}{2x} e^{x^2}\right) - \int \left(\frac{1}{2x} e^{x^2}\right)(3x^2) dx$

$\displaystyle \int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{2} \int (x)(e^{x^2}) dx $

$\displaystyle \int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{2} \left[(x)\left(\frac{1}{2x}e^{x^2}\right) - \int e^{x^2} dx\right] $

$\displaystyle \int (x^3)(e^{x^2}) dx = \frac{1}{2}x^{2}e^{x^2} - \frac{3}{4}e^{x^2} - \frac{1}{2x}e^{x^2} + c $

$\displaystyle \int (x^3)(e^{x^2}) dx = \frac{1}{2}e^{x^2}\left(x^2 - \frac{1}{x} - \frac{3}{2}\right) + c $

Now, Wolfram Alpha confirms the book's answer is correct.

So I decided to treat $\displaystyle \int e^{x^2} dx $ as a seperate integral and I've got:

$\displaystyle \frac{d}{dx} (e^{x^2}) = 2xe^{x^2} $

$\displaystyle \int e^{x^2} dx = \frac{1}{2x} e^{x^2} + c $

However, Wolfram Alpha says:

$\displaystyle \int e^{x^2} dx = \frac{1}{2} \sqrt{\pi}\ erfi(x) + c $

And I don't know why?