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Please solve this integration
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\[\int\limits_{0}^{1}lnxdx \div(1+lnx)^{2} = I ?\]
Where,(1+lnx) is +ve and lnx<0 for all x between 0 to 1 a(Upper limit)>0
ln means logx
Do you know that wolframalpha is pretty good at doing stuff like this..
Is the answer x/(1+log(x)) ?
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Well, no, because you have to put in the limits as well. There should be a numerical answer at the end
What would be the final answer?
http://www.wolframalpha.com/input/?i=integrate+%28ln+x%29%2F%281%2B+ln+x%29^2+from+0+to+1 integral does not converge
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