# How do you find the integral (x^4)(lnx)?

Apr 6, 2018

${x}^{3} - \frac{4}{3} \left({x}^{3}\right) + c$

#### Explanation:

Integration by parts is
$\int u \times \left(\mathrm{dv}\right) = u v - \int v \mathrm{du}$

Apply it to this integral:

$\int \left({x}^{4}\right) \left(I n x\right) \mathrm{dx}$

$u = {x}^{4}$
$\mathrm{dv} = I n x$
SO $v = \frac{1}{x}$

= ${x}^{4} \times \frac{1}{x} - \int \frac{1}{x} \times 4 {x}^{3} \mathrm{dx}$

= ${x}^{3} - 4 \int {x}^{2} \mathrm{dx}$

= ${x}^{3} - \frac{4}{3} \left({x}^{3}\right) + c$