Question #7e545

Sep 6, 2017

well, if we let $u = \ln x$ , then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

...so $\frac{1}{x} \mathrm{dx} = \mathrm{du}$

So, $\int {\left(\ln x\right)}^{3} / x \mathrm{dx}$

...can be re-written as $\int {u}^{3} \mathrm{du}$

$= {u}^{4} / 4 + C$

$= {\left(\ln x\right)}^{4} / 4 + C$