Integral of Logarithm

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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.

$\blacksquare$