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MEDAL - Integral: (lnx)^-2(x^-1) dx Does it need to be done by-parts?
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the equation editor would help not that i can do it, but at least i could read it
can we write that as \[\huge\rm \int\limits_{ }^{ } \frac{ 1 }{ (lnx)^2 \cdot x}\] \[(lnx)^{-2} = \frac{1}{(lnx)^2}\] samething x^{-1}=1/x
yes, sorry, I just didn't get to it quick enough
\[\int\limits_{ }^{} \frac{1}{x( lnx)^2} dx \] u-substitution
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ohhhh, sorry, it's late so I'm not thinking to the fullest ;)
thats fine but can you solve it from there ??
yes =)
(I took Calc I, II, III, and now I'm in DE)
okay nice
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ahhhhhhhhhh lol then you should help me :D
rusty nails don't help to well =D
ahh don't say that sir
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