# What is the antiderivative of xln(x) - x?

Jun 28, 2016

$= {x}^{2} / 2 \ln x - \frac{3}{4} {x}^{2} + C$

#### Explanation:

$\int \mathrm{dx} \setminus \left(x \ln x - x\right)$

for first term, $x \ln x$, we use IBP

$u = \ln x , u ' = \frac{1}{x}$
$v ' = x , v = {x}^{2} / 2$

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

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

for the second term

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

So the integral is

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

$= {x}^{2} / 2 \ln x - \frac{3}{4} {x}^{2} + C$