$\displaystyle \int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2}
{{{x^2}}}}}}}
{{{x^3}}}} dx
$
I'm not sure where to begin.
so I figured: $\displaystyle \frac{{ - 1}} {4}\int_{\ln e}^{\ln {e^3}} {{e^u}} du $ and $\displaystyle \frac{{ - 1}} {4}{e^u} = \frac{{ - {e^{\frac{2} {{{x^2}}}}}}} {4}_{\ln e}^{\ln {e^3}} = \frac{{ - {e^{\frac{2} {{{{(\ln {e^3})}^2}}}}}}} {4} - \frac{{ - {e^{\frac{2} {{{{(\ln e)}^2}}}}}}} {4} = \frac{{ - {e^{\frac{2} {9}}}}} {4} + \frac{{{e^2}}} {4} = 1.535 $
Hello, genlovesmusic09!
Note that: .$\displaystyle \ln(e) = 1,\;\;\ln(e^3) = 3 $$\displaystyle \int_{\ln e}^{\ln e^3} \frac{e^{\frac{2}{x^2}}}{x^3}\,dx$
We have: .$\displaystyle \int^3_1 e^{2x^{-2}}\left(x^{-3}\,dx\right) $
Let: .$\displaystyle u \:=\:2x^{-2} \quad\Rightarrow\quad du \:=\:-4x^{-3}dx \quad\Rightarrow\quad x^{-3}dx \:=\:-\tfrac{1}{4}du $
Substitute: .$\displaystyle \int e^u\left(-\tfrac{1}{4}\,du\right) \;=\;-\tfrac{1}{4}\int e^u\,du$
Got it?