Integral of Logarithm
Separate out into the two parts for $e$ and other bases
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Contents
Theorem
- $\displaystyle \int \ln x \ \mathrm d x = x \ln x - x + C$
- $\displaystyle \int \log_a x \ \mathrm d x = \frac 1 {\ln a}\left({x \ln x - x}\right) + C$
for $x > 0$.
Proof
Base $e$
- $\displaystyle \int \ln x \ \mathrm d x = \int 1 \cdot \ln x \ \mathrm d x$
From Integration by Parts:
- $\displaystyle \int fg' \, \mathrm dx = fg - \int f'g \, \mathrm dx$
Here:
| \(\displaystyle f\) | \(=\) | \(\displaystyle \ln x\) | |||||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle f'\) | \(=\) | \(\displaystyle \frac 1 x\) | Derivative of Natural Logarithm Function | |||||||||
| \(\displaystyle g'\) | \(=\) | \(\displaystyle 1\) | |||||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle g\) | \(=\) | \(\displaystyle x\) | Integral of Constant |
So:
| \(\displaystyle \int \ln x \ \mathrm d x\) | \(=\) | \(\displaystyle x \cdot \ln x -\int x \cdot \frac 1 x \ \mathrm dx\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x \ln x - x + C\) |
$\blacksquare$
Base $a$
Follows from Linear Combination of Integrals and Linear Operator on General Logarithm.
$\blacksquare$
