Different results when integrating 1/(xln(x)) partially/by substitution. By substitution I get ln(ln(x)). Partially something completely different: int 1/(x ln(x)) = int 1/x 1/(ln(x)) dx=(ln(x))/(ln(x)) - int -1/(x ln(x) ^2 )dx

Hallie Stanton

Hallie Stanton

Answered question

2022-11-04

Different results when integrating 1 / ( x ln ( x ) ) partially/by substitution.
By substitution I get l n ( l n ( x ) ). Partially something completely different:
1 x ln ( x ) = 1 x 1 ln ( x ) d x = ln ( x ) ln ( x ) 1 x ln ( x ) 2 d x
1 x ln ( x ) d x = 1 + C + 1 x ln ( x ) d x
Provided that this is correct, it doesn't really solve the integral - why is that?
Side info: this is not homework.

Answer & Explanation

Gilbert Petty

Gilbert Petty

Beginner2022-11-05Added 23 answers

This method will work for an integral of the form ( ln x ) r x d x if r 1, since then
u = ( ln x ) r , d v = 1 x d x gives
( ln x ) r x d x = ( ln x ) r + 1 r ( ln x ) r x d x , so
( 1 + r ) ( ln x ) r x d x = ( ln x ) r + 1 + C and so
( ln x ) r x d x = 1 r + 1 ( ln x ) r + 1 + C
(Of course, this is more easily found by substituting u = ln x.)
However, this gives no useful result when r = 1, and it does not lead to a contradiction since the integral is only determined up to a constant.

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