$\displaystyle \int_1^2 {{x^x}(1 + \ln x)} dx $ not sure where to begin
Follow Math Help Forum on Facebook and Google+
Originally Posted by genlovesmusic09 $\displaystyle \int_1^2 {{x^x}(1 + \ln x)} dx $ not sure where to begin Notice that if $\displaystyle u = x^x$ then $\displaystyle du = x^x(1 + \ln x)\,dx$
$\displaystyle \int_1^2 {du} = [{x^x}]_1^2 = 4 - 1 = 3 $ ?
yes.
View Tag Cloud