# What is the antiderivative of  x ln x ?

Apr 19, 2016

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

#### Explanation:

Using integration by parts:

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

In the case of

$\int x \ln x \mathrm{dx}$

We let

$u = \ln x \text{ "=>" "(du)/dx=1/x" "=>" } \mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\mathrm{dv} = x \mathrm{dx} \text{ "=>" "intdv=intxdx" "=>" } v = {x}^{2} / 2$

Thus, plugging these in, we see that

$\int x \ln x \mathrm{dx} = \left({x}^{2} / 2\right) \ln x - \int {x}^{2} / 2 \left(\frac{1}{x}\right) \mathrm{dx}$

$= \frac{{x}^{2} \ln x}{2} - \int \frac{x}{2} \mathrm{dx}$

$= \frac{{x}^{2} \ln x}{2} - {x}^{2} / 4 + C$