# How do you integrate (x^2 ln x) / x?

Jun 8, 2016

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

#### Explanation:

First note that one pair of $x$ terms will cancel:

$\frac{{x}^{2} \ln x}{x} = x \ln x$

So, we want to find:

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

We will use integration by parts, which takes the form:

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

So, for $\int x \ln x \mathrm{dx}$, let:

$\left\{\begin{matrix}u = \ln x \\ \mathrm{dv} = x \mathrm{dx}\end{matrix}\right.$

Differentiate $u$ and integrate $\mathrm{dv}$ to show that:

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

Plugging these back into the integration by parts formula:

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

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

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

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