$\displaystyle \int_1^2 {{x^x}(1 + \ln x)} dx

$

not sure where to begin

Printable View

- Oct 4th 2009, 09:35 AMgenlovesmusic09solving a integral with x to the x
$\displaystyle \int_1^2 {{x^x}(1 + \ln x)} dx

$

not sure where to begin - Oct 4th 2009, 09:38 AMJester
- Oct 4th 2009, 09:58 AMgenlovesmusic09
$\displaystyle \int_1^2 {du} = [{x^x}]_1^2 = 4 - 1 = 3

$

? - Oct 4th 2009, 11:26 AMKrizalid
yes.