# How do you integrate int x^4ln2x by integration by parts method?

Oct 18, 2016

$\int {x}^{4} \ln 2 x \mathrm{dx} = \frac{{x}^{5} \ln x}{5} - {x}^{4} / 4 + C$

#### Explanation:

Let $u \left(x\right) = \ln 2 x$ so $u ' \left(x\right) = \frac{2}{2 x} = \frac{1}{x}$
$v ' \left(x\right) = {x}^{4}$ so $v \left(x\right) = {x}^{5} / 5$
Integration by parts is
$\int u \left(x\right) v ' \left(x\right) \mathrm{dx} = u \left(x\right) v \left(x\right) - \int u ' \left(x\right) v \left(x\right) \mathrm{dx}$
Applying this we get
$\int {x}^{4} \ln 2 x \mathrm{dx} = \frac{{x}^{5} \ln x}{5} - \int {x}^{4} \frac{\mathrm{dx}}{x}$
$= \frac{{x}^{5} \ln x}{5} - \int {x}^{3} \mathrm{dx}$
$= \frac{{x}^{5} \ln x}{5} - {x}^{4} / 4 + C$