int_(1)^(oo) (ln(2x-1))/(x^2) dx Evaluate

Raina Gomez 2022-09-19 Answered
Evaluate 1 ln ( 2 x 1 ) x 2
My approach is to calc
1 X ln ( 2 x 1 ) x 2 d x
and then take the limit for the answer when X
However, I must do something wrong. The correct answer should be 2 ln ( 2 )
1 X ln ( 2 x 1 ) x 2 d x = [ 1 x ln ( 2 x 1 ) ] 1 X + 1 X 1 x × 2 2 x 1 d x = 1 X ln ( 2 X 1 ) + 2 1 X 1 x + 2 2 x 1 d x = 1 1 X ln ( 2 X 1 ) 2 ln X + 2 ln ( 2 X 1 )
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
1 X ln ( 2 X 1 ) 2 ln X + 2 ln ( 2 X 1 )
The first part will do to zero because of 1 X so we ignore that one, the second and third part:
2 ln X + 2 ln ( 2 X 1 ) = 2 ln ( 2 X 1 X ) = 2 ln ( 2 1 X ) = 2 ln ( 2 )
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Answers (2)

seguidora1e
Answered 2022-09-20 Author has 6 answers
Your derivative is incorrect, it should be
2 2 x 1
instead of
1 2 x 1
Everything else seems correct.
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Haiphongum
Answered 2022-09-21 Author has 2 answers
Another approach :
Setting x 1 x , we will obtain
1 ln ( 2 x 1 ) x 2   d x = 0 1 ln ( 2 x x ) d x = 0 1 [ ln ( 2 x ) ln x ]   d x .
Note that
ln y   d y = y   ln y y + C ,
hence
1 ln ( 2 x 1 ) x 2   d x = 2 ln 2.
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