How do you evaluate the definite integral #int (dx/(xsqrt(lnx)))# from #[1, e^6]#?

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Answer 1 · 2016-03-16T01:02:11

#2sqrt6#

Begin by using the substitution #u = ln(x)#.

From this we know that #du = 1/xdx#

Also the limits of our integral will change to:
#1: ln(1) = 0# and #e^6 : lne^6 = 6#

Thus we can substitute this into the integral as follows:

#int_1^(e^6)1/(xsqrt(lnx))dx=int_0^6 1/sqrt(u)du#

Now evaluate the integral:

#=int_0^6u^(-1/2)du=[2u^(1/2)]_0^6=2sqrt(6)#