How do you find the definite integral of #int 1/(xlnx)# from #[e,e^2]#?

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Answer 1 · 2017-01-24T06:40:18

The answer is #=ln2#

We do this integral by substitution

Let #u=lnx#, #=>#, #du=dx/x#

Therefore,

#intdx/(xlnx)=int(du)/(u)#

#=lnu=ln(lnx)#

So,

#int_e ^(e^2)dx/(xlnx)=[ln(lnx)]_e ^(e^2)#

#=(ln(lne^2)-ln(lne))#

#=ln2-ln1#

#=ln2#