pulled it out of my head... don't even know why, but will see if I can solve it.
\[\int\limits_{ }^{ } \ln(\ln x)~dx\]\[\int\limits_{ }^{ } \ln(\ln x) dx=x \ln(\ln x)- \int\limits_{ }^{ }\frac{1}{\ln x} \frac{1}{x} x~dx\]
you can see what I am doing,..
\[v=\ln(x)~~~~x=e^v,~~~~dv=\frac{1}{x}~dx\]
\[\int\limits_{ }^{ } \ln(\ln x)~dx=x \ln(\ln x) - \int\limits_{ }^{ }\frac{e^v}{v}~dv\]
\(\Large \int \dfrac{1}{\ln x} dx\) there is no answer to this integral in terms of standard functions
ohh, I did this before, it came out before in previous questions but it was just e^x/x dx, \[\int\limits_{ }^{ } \ln(\ln x)~dx=x \ln(\ln x) - \sum_{n=0}^{\infty} \frac{n!}{v^n}\]
\[\int\limits_{ }^{ } \ln(\ln x)~dx=x \ln(\ln x) - \sum_{n=0}^{\infty} \frac{n!}{(\ln x)^n}~\]
I have dealt 2 problems before with the e^x/x.
ofcourse you can use the series :) i was talking about standard functions like log f(x), e^{f(x)} , sin / cos /tan f(x), polynomials
oh no, series only.... good that x came to be e^v.
now though, I remember the integral of e^x/x .
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