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$$\int \frac{1}{x^3}e^{-x^2}dx$$

What I did?

put $1/x^2 = t; $

then $\int \frac{1}{x^3}e^{-x^2}dx$ will trasform into $\frac{-1}{2}\int e^{-1/t}dt$

I don't understand how to proceed there after.

EDIT:

This problem was given to me by my student. I inquired her, she corrected the problem as $\int \frac{1}{x^3}e^{-x^{-2}}dx$ that makes the problem very simple. Sorry for troubling you all.

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I am surprised this would be homework. There is no elementary function whose derivative is $\frac{1}{x^3}e^{-x^2}$. –  André Nicolas Oct 2 '12 at 15:47
    
@AndréNicolas: There is alos "Self-Learning" Tag. Its not a problem given to me in a class or something. I'm doing problems by self from a 1st year Bachelor's level book. –  claws Oct 2 '12 at 15:50
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You would have to quote the problem exactly. Numerical evaluation of related definite integrals would be possible, or a Fundamental Theorem of Calculus question, but not an indefinite integral question, unless they first define the exponential integral function –  André Nicolas Oct 2 '12 at 16:00

1 Answer 1

up vote 3 down vote accepted

You might try $u=x^2,\ du=2xdx$. Then $\int \frac{1}{x^3}e^{-x^2}dx=\frac 12\int\frac 1{u^2}e^{-u}du$, which can be integrated by parts using the Exponential integral, but not using more elementary functions.

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