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\[\int\limits_{ }^{ } \frac{1}{x~\ln x(\ln x~-1)}~dx\]
hint please?
Derivative of log is...
ln(x) ' = 1/x
substitute u = ln x
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du = 1/x dx
ok, then it becomes very good to solve, tnx
I see I see
:)
pellett clicked something
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\[\int\limits_{ }^{ } \frac{1}{u(u~-1)}~du\]
then, \[A(u)+B(u-1)=1\]
A=1 B=-1
\[\int\limits_{ }^{ } \frac{1}{u-1}-\frac{1}{u}~du\]
then, \[\color{red}{\ln( \ln x -1)-\ln(\ln x)+C}\]
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right?
noooooooo I missed it
you missed the prob, but did i miss anything?
9+10 ok
shut up, jk
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alright, I am out, will get suspended anyways, wouldn't?
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