# How do you evaluate the integral int xlnxdx?

Jan 10, 2017

$\frac{1}{4} {x}^{2} \left(2 \ln x - 1\right) + C$

#### Explanation:

This needs the method of 'integration by parts'. The formula of which needs to be known.

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

Care must be taken in the choice of $u$ & $\frac{\mathrm{dv}}{\mathrm{dx}}$ otherwise you end up with a more difficult, or impossible integration!

When logs are involved take the $u$ as the log part, the reason will become clear as we work through this problem.

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

$u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = x \implies v = \frac{1}{2} {x}^{2}$

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

$I = \frac{1}{2} {x}^{2} \ln x - \int \frac{1}{2} {x}^{2} \times \frac{1}{x} \mathrm{dx}$

$I = \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \int x \mathrm{dx}$

$I = \frac{1}{2} {x}^{2} \ln x - \frac{1}{4} {x}^{2} + C$

$I = \frac{1}{4} {x}^{2} \left(2 \ln x - 1\right) + C$