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Mathematics 12 Online
OpenStudy (anonymous):

what is the integral of 1/lnt dt

OpenStudy (anonymous):

I believe it is \[(t-1)/ln(t)\] however I am not a hundred percent sure. \[I = \int\limits(1/\ln(t))dt = t/\ln(t) + \int\limits(1/\ln^2(t))dt = t/\ln(t) - 1/\ln(t) = (t-1)/\ln(t)\]

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