Integral of Logarithm

 It has been suggested that this page or section be merged into Primitive of Logarithm of x. (Discuss)

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:

 $\displaystyle f$ $=$ $\displaystyle \ln x$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle f'$ $=$ $\displaystyle \frac 1 x$ $\quad$ Derivative of Natural Logarithm Function $\quad$ $\displaystyle g'$ $=$ $\displaystyle 1$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle g$ $=$ $\displaystyle x$ $\quad$ Integral of Constant $\quad$

So:

 $\displaystyle \int \ln x \ \mathrm d x$ $=$ $\displaystyle x \cdot \ln x -\int x \cdot \frac 1 x \ \mathrm dx$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle x \ln x - x + C$ $\quad$ $\quad$

$\blacksquare$

Base $a$

$\blacksquare$