# How to show that 1/(2a) ln |(x - a)/(x + a)| + C is equal to 1/(2a) ln |(x + a)/(x - a)| + K?

Jun 28, 2018

See below.

#### Explanation:

$\frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C = \frac{1}{2 a} \ln \left(| \frac{x + a}{x - a} {|}^{-} 1\right) + C$
$\frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C = - \frac{1}{2 a} \ln | \frac{x + a}{x - a} | + C$
$\frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C = \frac{1}{2 a} \ln | \frac{x + a}{x - a} | - \frac{1}{a} \ln | \frac{x + a}{x - a} | + C$
Let $K = - \frac{1}{a} \ln | \frac{x + a}{x - a} | + C$ so that the new constant $K$ "absorbs" the extra $- \frac{1}{a} \ln | \frac{x + a}{x - a} |$ along with $C$.
$\therefore \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C = \frac{1}{2 a} \ln | \frac{x + a}{x - a} | + K$

Jun 28, 2018

You can't snow it because it's not correct!

#### Explanation:

This is not possible. Suppose in fact that:

$\left(1\right) \text{ } \frac{1}{2 a} \ln \left\mid \frac{x - a}{x + a} \right\mid + C = \frac{1}{2 a} \ln \left\mid \frac{x + a}{x - a} \right\mid + K$

Since based on the properties of logarithms:

$\ln \left\mid \frac{x - a}{x + a} \right\mid = - \ln \left\mid \frac{x + a}{x - a} \right\mid$

it would follow that:

$\frac{1}{2 a} \ln \left\mid \frac{x - a}{x + a} \right\mid + C = - \frac{1}{2 a} \ln \left\mid \frac{x - a}{x + a} \right\mid + K$

and then:

$\frac{1}{a} \ln \left\mid \frac{x - a}{x + a} \right\mid = K - C$

But the function $\ln \left\mid \frac{x - a}{x + a} \right\mid$ is not constant.

graph{ln(((x-1)/(x+1))) [-40, 40, -20, 20]}

graph{ln(((x+1)/(x-1))) [-40, 40, -20, 20]}