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How do you evaluate the integral #int (ln(lnx))/x dx#?

Answer 1

Emily Ammerman

#lnx(ln(lnx)-1)+C#

First let #t=lnx#. This implies that #dt=1/xdx#. Then:

#intln(lnx)/xdx=intln(lnx)1/xdx=intln(t)dt#

Now we can use integration by parts which takes the form #intudv=uv-intvdu#. Let:

#{(u=lnt" "=>" "du=1/tdt),(dv=dt" "=>" "v=t):}#

Then:

#=intln(t)dt=tlnt-intt1/tdt#

#=tlnt-intdt#

#=tlnt-t#

#=t(lnt-1)#

#=lnx(ln(lnx)-1)+C#

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Answer 2

Cameron Alexander

To evaluate the integral (\int \frac{\ln(\ln x)}{x} , dx), you can use integration by parts, setting (u = \ln(\ln x)) and (dv = \frac{1}{x} , dx). Then, find (du) and integrate (dv) to obtain (v). Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Once you've calculated the value of the integral, don't forget to add the constant of integration, (C).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.