# integral with lne

• Oct 4th 2009, 10:22 AM
genlovesmusic09
integral with lne
$\int_{\ln e}^{\ln {e^3}} {\frac{{{e^{\frac{2}
{{{x^2}}}}}}}
{{{x^3}}}} dx
$

I'm not sure where to begin.
• Oct 4th 2009, 10:42 AM
TheEmptySet
Quote:

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.
• Oct 4th 2009, 10:50 AM
genlovesmusic09
would this be right?
$\frac{{ - 1}}
{4}\int_{\ln e}^{\ln {e^3}} { - 4{e^u}} du
$
• Oct 4th 2009, 07:15 PM
genlovesmusic09
so I figured: $\frac{{ - 1}} {4}\int_{\ln e}^{\ln {e^3}} {{e^u}} du$ and $\frac{{ - 1}} {4}{e^u} = \frac{{ - {e^{\frac{2} {{{x^2}}}}}}} {4}_{\ln e}^{\ln {e^3}} = \frac{{ - {e^{\frac{2} {{{{(\ln {e^3})}^2}}}}}}} {4} - \frac{{ - {e^{\frac{2} {{{{(\ln e)}^2}}}}}}} {4} = \frac{{ - {e^{\frac{2} {9}}}}} {4} + \frac{{{e^2}}} {4} = 1.535$
• Oct 4th 2009, 10:52 PM
Soroban
Hello, genlovesmusic09!

Quote:

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