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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\ln x$: $14.529$