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ParthKohli
Group Title
\[\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1x}\right)^2 \mathrm{dx}\] @Mimi_x3
 one year ago
 one year ago
ParthKohli Group Title
\[\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1x}\right)^2 \mathrm{dx}\] @Mimi_x3
 one year ago
 one year ago

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Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
have fun mimi =]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
could work backwards in figuring out what f(x) is
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
rather than complete the integral itself
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
lol, looks like those improper integrals?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
using what we know if of this \[\int f(x)dx=F(x)\]
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
@Outkast3r09 i dont think that would work
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
can not be done by elementary terms.. http://www.wolframalpha.com/input/?i=integrate+%28+%28%281%2Flnx%29%2B%281%2F%281x%29%29%29^2
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
and this is why i hardly ever answer parth questions
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
This is not my question; I found it from somewhere else. :)
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
well, i will try..will take a while tho lol
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I don't even get what the solution says. Hehe :D
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Only legit method I see is using Simpson's approximation formula.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
seriously i don't know ... but since you say it, then my guess must be correct :D
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Hoho ;)
 one year ago
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