Primitive of Logarithm of x over x squared

Theorem

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

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

Thus: