How do you integrate #1/(xlnx)dx#?

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Feb 15, 2015

Answer : # ln ln x + c #

Details :

Basic Fact used:
This is an application of the fact that # int dx/x = ln x + c #

Substitution :
Look at the form of the integral: # 1/ln x dx/x #. Remembering the fact above, one can substitute # t = ln x # so that # dt = dx /x #.

As a result the integral simplifies to #int 1/ln x dx/x = int dt/t = ln t + c#

Back Substitute #t = ln x# to get the answer.

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vince Share
Feb 15, 2015

Hello !

I propose another solution.

Remember that #(\ln(u))' = \frac{u'}{u}# if #u# is a positive differentiable function.

Take #u (x) = \ln(x)# for #x>1# : it's a positive differentiable function.

Remark that #\frac{u'(x)}{u(x)} = \frac{\frac{1}{x}}{\ln(x)} = \frac{1}{x\ln(x)}#, then

#\int \frac{\text{d}x}{x\ln(x)} = \ln(u(x)) + c = \ln(\ln(x)) + c#,

where #c# is a real constant.

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