# 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

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

Thus:

 $\ds \int \frac {\ln x} {x^2} \rd x$ $=$ $\ds \frac {x^{-1} } {-1} \paren {\ln x - \frac 1 {-1} } + C$ Primitive of $x^m \ln x$, setting $m = -2$ $\ds$ $=$ $\ds \frac {-\ln x} x - \frac 1 x + C$ simplifying

$\blacksquare$