# How do I find the integral intx^5*ln(x)dx ?

Sep 24, 2014

By Integration by Parts,

$\int {x}^{5} \ln x \mathrm{dx} = {x}^{6} / 36 \left(6 \ln x - 1\right) + C$

Let us look at some details.

Let $u = \ln x$ and $\mathrm{dv} = {x}^{5} \mathrm{dx}$.
$R i g h t a r r o w \mathrm{du} = \frac{\mathrm{dx}}{x}$ and $v = {x}^{6} / 6$

By Integration by Parts

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$,

we have

$\int \left(\ln x\right) \cdot {x}^{5} \mathrm{dx} = \left(\ln x\right) \cdot {x}^{6} / 6 - \int {x}^{6} / 6 \cdot \frac{\mathrm{dx}}{x}$

by simplifying a bit,

$= {x}^{6} / 6 \ln x - \int {x}^{5} / 6 \mathrm{dx}$

by Power Rule,

$= {x}^{6} / 6 \ln x - {x}^{6} / 36 + C$

by factoring out ${x}^{6} / 36$,

$= {x}^{6} / 36 \left(6 \ln x - 1\right) + C$