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

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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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