Question

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

Answer 1

See below.

Explanation:

`1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln(|(x+a)/(x-a)|^-1)+C`
`1/(2a)ln|(x-a)/(x+a)|+C=-1/(2a)ln|(x+a)/(x-a)|+C`
`1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln|(x+a)/(x-a)|-1/aln|(x+a)/(x-a)|+C`
Let `K=-1/aln|(x+a)/(x-a)|+C` so that the new constant `K` "absorbs" the extra `-1/aln|(x+a)/(x-a)|` along with `C`.
`:.1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln|(x+a)/(x-a)|+K`

Answer 2

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

Explanation:

This is not possible. Suppose in fact that:

`(1) " "1/(2a)ln abs ((x-a)/(x+a)) +C = 1/(2a)ln abs ((x+a)/(x-a))+K`

Since based on the properties of logarithms:

`ln abs ((x-a)/(x+a)) = -ln abs ((x+a)/(x-a))`

it would follow that:

`1/(2a)ln abs ((x-a)/(x+a)) +C = -1/(2a)ln abs ((x-a)/(x+a))+K`

and then:

`1/a ln abs ((x-a)/(x+a)) = K-C`

But the function `ln abs ((x-a)/(x+a))` is not constant.

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

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