Answer
$\frac{1}{2}w^2\ln w-\frac{1}{4}w^2+C$
Work Step by Step
Consider $u=\ln w$ and $dv=w dw$.
Then, $du=\frac{1}{w} dw$ and $v=\frac{1}{2}w^2$.
Using the Integration by Parts,
$\int w\ln w dw=\int \ln w (w dw)$
$=(\ln w)\cdot \frac{1}{2}w^2-\int \frac{1}{2}w^2\cdot \frac{1}{w}dw$
$=\frac{1}{2}w^2\ln w-\int \frac{1}{2}wdw$
$=\frac{1}{2}w^2\ln w-\frac{1}{4}w^2+C$
