X xc630 Junior Member Joined Sep 1, 2005 Messages 164 Dec 4, 2007 #1 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
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
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Dec 4, 2007 #2 Re: substitution rule Hello, xc630! \(\displaystyle \int \frac{e^{\frac{1}{x}}}{x^2}\,dx\) Click to expand... \(\displaystyle \text{Note that we have: }\;\int e^{x^{\text{-}1}}\left(x^{\text{-}2}\right)\,dx\) \(\displaystyle \text{Now let: }\:u \:=\:x^{\text{-}1}}\quad\Rightarrow\quad du \:=\:-x^{\text{-}2}dx\quad\Rightarrow\quad x^{\text{-}2}dx \:=\:-du\) \(\displaystyle \text{Substitute: }\;\int e^u (-du) \;=\;-\int e^u du\) . . . Got it?
Re: substitution rule Hello, xc630! \(\displaystyle \int \frac{e^{\frac{1}{x}}}{x^2}\,dx\) Click to expand... \(\displaystyle \text{Note that we have: }\;\int e^{x^{\text{-}1}}\left(x^{\text{-}2}\right)\,dx\) \(\displaystyle \text{Now let: }\:u \:=\:x^{\text{-}1}}\quad\Rightarrow\quad du \:=\:-x^{\text{-}2}dx\quad\Rightarrow\quad x^{\text{-}2}dx \:=\:-du\) \(\displaystyle \text{Substitute: }\;\int e^u (-du) \;=\;-\int e^u du\) . . . Got it?
X xc630 Junior Member Joined Sep 1, 2005 Messages 164 Dec 4, 2007 #3 Re: substitution rule I didn't think of bringing up the x^2. Thanks soroban!
