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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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    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 function2012-10-02

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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.