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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#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#
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]}
