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I need help with evaluating the integral:

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


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What do you call to solve an integral? – Did Apr 13 '12 at 10:18
it could not be solved using elementary functions – dato datuashvili Apr 13 '12 at 10:29
You could relate it to an incomplete Gamma-function. And this function has been well studied. – Raskolnikov Apr 13 '12 at 10:37
If it were a definite integral, it could be accurately approximated. If it were $\int x^2e^{-x^3}dx$, the antiderivative can be expressed in terms of elementary functions. But for this one, it cannot. See this post covering a relevant theorem of Liouville and the Risch algorithm for more info. If this is a homework problem, it is an error. What is the precise problem or your real need? – bgins Apr 13 '12 at 10:58

Typo perhaps? If it is meant to be either $\int x^2 e^{x^{3}}dx$ or $\int x e^{x^{2}} dx$ then these can be evaluated very simply.

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Yes, it was typo in the homework. It was supposed to be $x^2 e^{x^3}dx$. Thanks anyway. – aldo Apr 13 '12 at 11:28
No problem, glad to be of help. I take it you're good from now on? It should be obvious since I assume the very integral in question is being evaluated as you had most probably been studying "integration by inspection". – Autolatry Apr 13 '12 at 11:36
@aldo: You might want to edit your question to account for this new development... – J. M. Apr 15 '12 at 16:39
Take t = x^3 
dt = 3x^2 dx and x = t^(1/3)

and substitute x and find integration by using LIATE

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We get $(1/3)\int e^{-t} t^{-1/3}\,dt$, unfortunately still not an "elementary" integral. – GEdgar Apr 13 '12 at 13:05
Its look like ∫f(x)g(x)dx. – Prasad G Apr 13 '12 at 13:35
Its look like ∫f(t)g(t)dt. Take f(t)= e^(-t) and g(t) = t^(-1/3) – Prasad G Apr 13 '12 at 13:47
Doesn't help... – GEdgar Apr 13 '12 at 16:35

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