The Natural Logarithmic Function: Differentiation and Integration

The natural logarithm, generally written as $\ln(x)$, is the logarithm with the base e, where e is an irrational and transcendental constant approximately equal to $2.718281828$.

The derivative of the natural logarithm is given by:

$\dfrac{d}{dx}\ln(x) = \dfrac{1}{x}$

This leads to the Taylor series for $\ln(1 + x)$ around $0$: 

$\ln(1+x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \cdots$ 

for $\left | x \right | \leq 1$ (unless $x = -1$).

Taylor Series Approximations for $\ln(1+x)$

The Taylor polynomials for $\ln(1 + x)$ only provide accurate approximations in the range $-1 < x \leq 1$. Note that, for $x>1$, the Taylor polynomials of higher degree are worse approximations.

Substituting $x − 1$ for $x$, we obtain an alternative form for $\ln(x)$ itself: 

$\ln(x) = (x - 1) - \dfrac{(x - 1)^{2}}{2} + \dfrac{(x - 1)^{3}}{3} - \cdots$ 

for $\left | x -1 \right | \leq 1$ (unless $x = 0$).

By using Euler transform, we reach the following equation, which is valid for any $x$ with absolute value greater than $1$: 

$\ln\dfrac{x}{x-1} = \dfrac{1}{x} + \dfrac{1}{2x^{2}} + \dfrac{1}{3x^{3}} + ...$

The natural logarithm allows simple integration of functions of the form $g(x) = \frac{f '(x)}{f(x)}$: an antiderivative of $g(x)$ is given by $\ln\left(\left|f(x)\right|\right)$. This is the case because of the chain rule and the following fact: 

$\dfrac{d}{dx}\left(\ln\left | x \right |\right) = \dfrac{1}{x}$

In other words:

$\displaystyle{\int \dfrac{1}{x}dx = \ln\left | x \right | + C}$

and 

$\int \frac{f'(x)}{f(x)}dx = ln\left | f(x) \right | + C$

Here is an example in the case of $g(x) = \tan(x)$:

$\displaystyle{\int \tan (x)dx = \int \frac{\sin (x)}{\cos (x)}dx}$

$\displaystyle{\int \tan (x)dx = \int \frac{\frac{-d}{dx}cos (x)}{\cos (x)}dx}$

Letting $f(x) = \cos(x)$ and $f'(x)= – \sin(x)$:

$\displaystyle{\int \tan (x)dx = -\ln\left | \cos(x) \right | + C}$

where $C$ is an arbitrary constant of integration.

The natural logarithm can be integrated using integration by parts: 

$\displaystyle{\int \ln(x)dx = x\ln(x) - x + C}$