integral with lne

**genlovesmusic09** · October 4th 2009, 09:22 AM

$\int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2} {{{x^2}}}}}}} {{{x^3}}}} dx $

I'm not sure where to begin.

**TheEmptySet** · October 4th 2009, 09:42 AM

Originally Posted by genlovesmusic09

$\int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2} {{{x^2}}}}}}} {{{x^3}}}} dx $

I'm not sure where to begin.

It is a u sub

$u=\frac{2}{x^2} \implies du=-\frac{4}{x^3}dx$

This should get you started good luck.

**genlovesmusic09** · October 4th 2009, 09:50 AM

would this be right?
$\frac{{ - 1}} {4}\int_{\ln e}^{\ln {e^3}} { - 4{e^u}} du $

**genlovesmusic09** · October 4th 2009, 06:15 PM

so I figured: $\frac{{ - 1}} {4}\int_{\ln e}^{\ln {e^3}} {{e^u}} du$ and $\frac{{ - 1}} {4}{e^u} = \frac{{ - {e^{\frac{2} {{{x^2}}}}}}} {4}_{\ln e}^{\ln {e^3}} = \frac{{ - {e^{\frac{2} {{{{(\ln {e^3})}^2}}}}}}} {4} - \frac{{ - {e^{\frac{2} {{{{(\ln e)}^2}}}}}}} {4} = \frac{{ - {e^{\frac{2} {9}}}}} {4} + \frac{{{e^2}}} {4} = 1.535$

**Soroban** · October 4th 2009, 09:52 PM

Hello, genlovesmusic09!

$\int_{\ln e}^{\ln e^3} \frac{e^{\frac{2}{x^2}}}{x^3}\,dx$

Note that: . $\ln(e) = 1,\;\;\ln(e^3) = 3$

We have: . $\int^3_1 e^{2x^{-2}}\left(x^{-3}\,dx\right)$

Let: . $u \:=\:2x^{-2} \quad\Rightarrow\quad du \:=\:-4x^{-3}dx \quad\Rightarrow\quad x^{-3}dx \:=\:-\tfrac{1}{4}du$

Substitute: . $\int e^u\left(-\tfrac{1}{4}\,du\right) \;=\;-\tfrac{1}{4}\int e^u\,du$

Got it?

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