anonymous
  • anonymous
integral from 1 to 2 1/(x(lnx))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{1}^{2} 1/(xln(x)) dx\] \[\int\limits_{1}^{2} (1/x) (1/ln(x)) dx\] \[u=\ln(x)\] \[du=(1/x)dx\] \[\int\limits_{0}^{ln(2)} (1/u) du\] \[\int\limits_{0}^{ln(2)}u ^{-1}du\] Think you got it from there?
Callisto
  • Callisto
I found something weird... |dw:1332397771190:dw|
experimentX
  • experimentX
let ln(x) = u, then 1/x*dx = du => integral 1/u*du => ln(u) substitute the value of u, you have ln(ln(x)) and evaluate it

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anonymous
  • anonymous
Oh yikes! The last step on mine is unnecessary - and possibly confusing!
anonymous
  • anonymous
what if we dont want to change our b and a terms and want to keep them as they are
anonymous
  • anonymous
Then after you integrate, you would substitute back what you started with for "u"
Callisto
  • Callisto
i didn't change the variable.. did i ?
anonymous
  • anonymous
so what would the final answer be?
anonymous
  • anonymous
I did ::hides face::
anonymous
  • anonymous
Callisto and ExperimentX both have it there.
Callisto
  • Callisto
ln|ln2|
anonymous
  • anonymous
thanks!!! you guys are the best
Callisto
  • Callisto
i'm not sure if my answer is correct. i've been making too many mistakes these days. But have you learnt how to do it?
anonymous
  • anonymous
no, not fully
Callisto
  • Callisto
which part you don't understand?
anonymous
  • anonymous
i do understand now thank you. i was just wondering if ln(ln2) can be reduced even more
anonymous
  • anonymous
\[\int\limits_{}^{}(1/u)du\] ln(u)] Then sub back \[\ln(\ln(x))]_{1}^{2}\] Then evaluate ln(ln(2))-ln(ln(1)) Wait a minute... ln(ln(2))-ln(0) ln(ln(2))-undefined?
anonymous
  • anonymous
Ha! I'm sorry - I'm making this question about me. So because the second part of the expression is undefined we just ignore it and go with ln(ln(2)) as the answer?
anonymous
  • anonymous
Wait - I don't think you can integrate this one because this function is not continuous from [a,b]
anonymous
  • anonymous
i dont know... :(
anonymous
  • anonymous
so what now? improper integrals? can that be done or no?
anonymous
  • anonymous
Yeah, it would be an improper integral because the limit of 1 is a problem.
anonymous
  • anonymous
I believe it can be done - it's just not as easy as we first suspected. :)
anonymous
  • anonymous
please continue then because i still cannot get an answer lol
anonymous
  • anonymous
I'll do my best - I'm looking it up right now. :) It's been a while since I played with these.
Callisto
  • Callisto
this time we need wolf to help http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B1%7D%5E%7B2%7D+1%2F%28xln%28x%29%29+dx&dataset=&equal=Submit
Callisto
  • Callisto
it's really my mistake sorry :(
anonymous
  • anonymous
but wolf ram gave us no steps?
anonymous
  • anonymous
I've got an example for you to look at:
1 Attachment
Callisto
  • Callisto
as long as it comes to limit, i can't help anymore.. sorry!
anonymous
  • anonymous
\[\int\limits_{1}^{2}1/(x(\ln(x))dx\] but, because there is a discontinuity at the left endpoint: \[\int\limits\limits_{1}^{2}1/(x(\ln(x))dx=\lim_{t \rightarrow 1^+}\int\limits\limits_{t}^{2}1/(x(\ln(x))dx\]
anonymous
  • anonymous
So then, we can integrate the way that we did before.
anonymous
  • anonymous
\[=\lim_{t \rightarrow 1^+}\int\limits_{t}^{2}1/(x(\ln(x))dx\] \[u = \ln(x)\] \[du=(1/x)dx\] \[\lim_{t \rightarrow 1^+}\int\limits_{}^{}(1/u)du\] \[\lim_{t \rightarrow 1^+}\ln(\ln(x))_{t}^{2}\] \[\lim_{t \rightarrow 1^+}\ln(\ln(2))-\ln(\ln(t))\] The problem here is that we are still left undefined with that second piece. I wish I could help more. I'm wondering if you have to use integration by parts and possibly L'Hospital? Good luck, friend!
myininaya
  • myininaya
Hey @TheFigure I was lurking a bit and seen something you posted and I just wanted to tell you. Since that one part doesn't converge, then the whole thing does not converge. :)
myininaya
  • myininaya
No actually I seen one of @callisto 's post. My bad. lol
Callisto
  • Callisto
I know.. :(

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