Question
Evaluate: $\int x \log x d x$
$\int u d v=u v-\int v d u$
Choosing u = logx and dv = xdx
$d u=\frac{1}{x} d x$
$v=\frac{x^2}{2}$
$\therefore \int x \log x d x=\log x \frac{x^2}{2}-\int \frac{x^2}{2} \frac{1}{x} d x$
$=\frac{x^2}{2} \log x-\frac{1}{2} \int x d x$
$=\frac{x^2}{2} \log x-\frac{1}{2} \frac{x^2}{2}+C$
$=\frac{x^2}{2} \log x-\frac{x^2}{4}+C$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.