Primitive of x by Logarithm of x/Proof 2

Theorem

 * $\displaystyle \int x \ln x \ \mathrm d x = \frac {x^2} 2 \left({\ln x - \frac 1 2}\right) + C$

Proof
From Primitive of $x^m \ln x$:
 * $\displaystyle \int x^m \ln x \ \mathrm d x = \frac {x^{m + 1} } {m + 1} \left({\ln x - \frac 1 {m + 1} }\right) + C$

The result follows by setting $m = 1$.