# What is the antiderivative of (x^4)(lnx)?

Apr 7, 2016

$\frac{1}{5} {x}^{5} \ln x - \frac{1}{25} {x}^{5} + C$

#### Explanation:

Note that:

$\frac{d}{\mathrm{dx}} \frac{1}{5} {x}^{5} \ln x = {x}^{4} \ln x + \frac{1}{5} {x}^{5} \frac{1}{x} = {x}^{4} \ln x + \frac{1}{5} {x}^{4}$

$\frac{d}{\mathrm{dx}} \frac{1}{25} {x}^{5} = \frac{1}{5} {x}^{4}$

So:

$\frac{d}{\mathrm{dx}} \left(\frac{1}{5} {x}^{5} \ln x - \frac{1}{25} {x}^{5}\right) = {x}^{4} \ln x$

So the antiderivative of ${x}^{4} \ln x$ is $\frac{1}{5} {x}^{5} \ln x - \frac{1}{25} {x}^{5} + C$