Question

Integrate the function $x\log x$

Answer 1

Let $I=\int x\log xdx$
Taking $\log x$ as first function and $x$ as second function and using integrating by parts, we obtain
$I=\log x\int xdx-\int \{\left(\frac{d}{dx}\log x\right)\int xdx\}dx$
$=\log x\cdot \frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx$
$=\frac{x^2\log x}{2}-\int \frac{x}{2}dx$
$=\frac{x^2\log x}{2}-\frac{x^2}{4}+C$