Primitive of Logarithm of x over x squared

Theorem

 * $\ds \int \frac {\ln x} {x^2} \rd x = \frac {-\ln x} x - \frac 1 x + C$

Proof
From Primitive of $x^m \ln x$:
 * $\ds \int x^m \ln x \rd x = \frac {x^{m + 1} } {m + 1} \paren {\ln x - \frac 1 {m + 1} } + C$

Thus: