# Integral of Logarithm

Jump to navigation
Jump to search

## 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:

\(\ds f\) | \(=\) | \(\ds \ln x\) | ||||||||||||

\(\ds \implies \ \ \) | \(\ds f'\) | \(=\) | \(\ds \frac 1 x\) | Derivative of Natural Logarithm Function | ||||||||||

\(\ds g'\) | \(=\) | \(\ds 1\) | ||||||||||||

\(\ds \implies \ \ \) | \(\ds g\) | \(=\) | \(\ds x\) | Integral of Constant |

So:

\(\ds \int \ln x \ \mathrm d x\) | \(=\) | \(\ds x \cdot \ln x -\int x \cdot \frac 1 x \ \mathrm dx\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds x \ln x - x + C\) |

$\blacksquare$

### Base $a$

Follows from Linear Combination of Integrals and Linear Operator on General Logarithm.

$\blacksquare$