I'm lost on this one, just need a starting point.
$\displaystyle \int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$
Thanks for any help!
Consider $\displaystyle \int 6e^{3x}~dx$
Let $\displaystyle u = 3x$
$\displaystyle \Rightarrow du = 3~dx$
$\displaystyle \Rightarrow \frac 13 ~du = dx$
$\displaystyle \int 6e^{{\color{red}3x}}~{\color{blue}dx} = \int 6e^{\color{red}u} \cdot {\color{blue}\frac 13~du}$
$\displaystyle = 2 \int e^u~du$