Question about $\int e^{x^2} dx.$ Why can't it be integrated using standard methods? I have a question about $\int e^{x^2} dx.$
I understand that if you try and use u-substitution, you will end up with an integral that has both u and x in the integrand, which cannot be evaluated.
However, there is a "shortcut method" of the reverse chain rule by saying that since d/du (e^u) = (e^u)*u', therefore the integral of e^u du is equal to (e^u)/u'. This rule works in most cases, such as with e^(3x), for example.
If I were to use this "shortcut" version of the reverse chain rule, we could then say that the integral of e^(x^2) dx = e^(x^2)/(2x) + c (if doing an indefinite integral). We could verify that this is correct by taking the derivative of our answer and sure enough, it takes us back to where we started.
However, if I evaluate this as a definite integral between two bounds by hand, I get a different result than if I evaluate it using my TI-84 calculator. Therefore, I assume the integral I calculated earlier must be incorrect. Why? I'm hoping someone can help me understand. And since it is not correct, why can I take the derivative of the result and it takes me back to where I started? Usually, this is a good check to see if you executed the integral properly.
I appreciate any help and thanks in advance!
 A: You say that $$\frac{d}{du}e^u=u'e^u.$$ This is not correct, because in reality $$\frac{d}{du}e^u=e^u$$ because the variable of differentiation is $u$. The correct statement is rather $$\frac{d}{dx}e^u=u'e^u$$ where $u=u(x)$ should be interpreted as a function of $x$. Therefore your "shortcut" version of the reverse chain rule, $$\int e^u du = \frac{e^u}{u'}$$ is false and I think represents that you are confusing your variables $u$ and $x$.
In particular your claim that $$\int e^{x^2}dx=\frac{e^{x^2}}{2x}+C$$ is also wrong: no easy elementary expression for the integral exists. If you differentiate the RHS:
$$\frac{d}{dx}\frac{e^{x^2}}{2x}=\frac12\frac{d}{dx}x^{-1}e^{x^2}=\frac12\left(-x^{-2}e^{x^2}+x^{-1}\cdot2xe^{x^2}\right)\neq e^{x^2}.$$
Your mistake is forgetting to use the product rule and forgetting the first term.
A: You have it in reverse.
If $\frac{d}{dx}e^u = u'e^u$ then $\int u'e^udx = e^u$ where $u = u(x)$
If $u(x) = cx$ where c is a constant.
$\frac{d}{dx}e^{cx} = ce^{cx}$
Taking the integral on both sides
$\int \frac{d}{dx}e^{cx}dx = \int ce^{cx}dx$
$e^cx = \int ce^{cx}dx = c\int e^{cx}dx$
$\Rightarrow \frac{e^cx}{c} = \int e^{cx}dx$
$\Rightarrow \int e^{cx}dx = \frac{e^{cx}}{c}$
Notice, that is not $c'$. That's a constant. Note that $c' = 0$ for any constant.
As Yifan mentioned, you may be confusing your variables $u$ and $x$ and not differentiating properly. Use the product rule/quotient rule to differentiate this.
Another way you can know that your antiderivative is wrong:
Consider the interval $[0,1]$. $f(x) = e^{x^2}$ is continuous on this interval. By the FTC, if you have an antiderivative $F(x)$, where $F'(x) = f(x)$ then
$$\int_0^1 f(x) = F(1) - F(0)$$
but $F(0)$ is undefined in the antiderivative $F(x) = \frac{e^{x^2}}{2x}$ you found , and even if you take the limit, it does not exist. Hence your antiderivative must be wrong.
A: Look up "Integration in finite terms."
This will show why $e^{x^2}$
cannot be integrated as you want.
