# How do you integrate (x^4)(lnx)?

Jul 26, 2016

$\int \ln x \times {x}^{4} \mathrm{dx} = {x}^{5} / 5 \left(\ln x - \frac{1}{5}\right)$

#### Explanation:

WE can use integration by parts $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = \ln x$ and $v = {x}^{5} / 5$

Hence $\mathrm{du} = \frac{\mathrm{dx}}{x}$ and $\mathrm{dv} = {x}^{4} \mathrm{dx}$ and $\int u \mathrm{dv} = u v = \int v \mathrm{du}$ is

$\int \ln x \times {x}^{4} \mathrm{dx} = \int u \mathrm{dv} = u v - \int v \mathrm{du}$

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

= $\frac{\ln x \times {x}^{5}}{5} - \frac{1}{5} \int {x}^{4} \mathrm{dx}$

= $\frac{\ln x \times {x}^{5}}{5} - {x}^{5} / 25$

= ${x}^{5} / 5 \left(\ln x - \frac{1}{5}\right)$