substitution rule: integral of [(e^(1/x)) / (x^2)] dx

[xc630]

Hello,

I would like help with evaluating the integral of [(e^(1/x)) / (x^2)] dx

I'm not sure what I should let u=
I've tried x^2 and e^ (1/x) but don't seem to be getting anywhere.

Thanks

 

[soroban]

Re: substitution rule

Hello, xc630!

\(\displaystyle \int \frac{e^{\frac{1}{x}}}{x^2}\,dx\)

\(\displaystyle \text{Note that we have: }\;\int e^{x^{\text{-}1}}\left(x^{\text{-}2}\right)\,dx\)

\(\displaystyle \text{Now let: }\:u \:=\:x^{\text{-}1}}\quad\Rightarrow\quad du \:=\:-x^{\text{-}2}dx\quad\Rightarrow\quad x^{\text{-}2}dx \:=\:-du\)

\(\displaystyle \text{Substitute: }\;\int e^u (-du) \;=\;-\int e^u du\)

. . . Got it?

 

[xc630]

Re: substitution rule

I didn't think of bringing up the x^2. Thanks soroban!