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

Nov 23, 2016

$\frac{1}{2} {x}^{2} \ln x - \frac{3}{4} {x}^{2} + C$

#### Explanation:

This is the same as asking

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

Splitting up the integral:

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

The second can be integrated using the power rule for integration:

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

For the remaining integral, use integration by parts. This takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. For $\int x \ln x \mathrm{dx}$, let:

$\left\{\begin{matrix}u = \ln x & \implies & \mathrm{du} = \frac{1}{x} \mathrm{dx} \\ \mathrm{dv} = x \mathrm{dx} & \implies & v = {x}^{2} / 2\end{matrix}\right.$

Thus:

$= u v - \int v \mathrm{du} - {x}^{2} / 2$

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

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

$= \frac{1}{2} {x}^{2} \ln x - \frac{3}{4} {x}^{2} + C$