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

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Answer 1 · 2018-06-28T14:20:44

See below.

Answer 2 · 2018-06-28T15:09:49

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

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]}