# What is the antiderivative of xlnx?

Sep 21, 2016

$\int x \ln x \mathrm{dx} = {x}^{2} / 4 \left(2 \ln x - 1\right) + C , \mathmr{and} , {x}^{2} / 4 \ln \left({x}^{2} / e\right) + C , \mathmr{and} \ln {\left({x}^{2} / e\right)}^{{x}^{2} / 4} + C .$

#### Explanation:

Let $I = \int x \ln x \mathrm{dx}$

We use the following Rule of Integration by Parts (IBP) :

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left[\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right] \mathrm{dx} .$ We take,

$u = \ln x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} , \mathmr{and} , v = x \Rightarrow \int v \mathrm{dx} = \int x \mathrm{dx} = {x}^{2} / 2$.

Hence,

$I = {x}^{2} / 2 \ln x - \int \left[\frac{1}{x} \cdot {x}^{2} / 2\right] \mathrm{dx}$

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

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

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

$= {x}^{2} / 4 \left(2 \ln x - 1\right) ,$ or,

$= {x}^{2} / 4 \left(\ln {x}^{2} - \ln e\right)$

$= {x}^{2} / 4 \ln \left({x}^{2} / e\right)$, or,

$= \ln {\left({x}^{2} / e\right)}^{{x}^{2} / 4} .$

Enjoy Maths.!