Primitive of Power of x by Logarithm of x

Theorem

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

where $m \ne -1$.

Proof
With a view to expressing the primitive in the form:
 * $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

Then:

Also see

 * Primitive of $\dfrac {\ln x} x$ for the case where $m = -1$.