Primitive of Logarithm of x over x squared

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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:

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

$\blacksquare$


Sources