PhoenixFire
  • PhoenixFire
\[\int_{e^-1}^{e} {\frac{1}{t(1+\left | ln(t) \right |)} dt}\] Can anyone guide me on starting this integral? I've tried numerous times but keep getting a division by zero. No need to answer the full question, just a hint on starting would be helpful.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
tell me what have you done untill now?
hartnn
  • hartnn
substitute 1+ln t = x hace u donee this ?
hartnn
  • hartnn
**have done

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hartnn
  • hartnn
and if u have done that, have you changed the limits ?
PhoenixFire
  • PhoenixFire
I have tried x=1+ln(t), the limits become 2 and 0. when you integrate 1/x you get lnx.. but you can't put in the limit of 0 into that. this is what I kept getting stuck on.
anonymous
  • anonymous
then take the limits of 0... that what we do..
hartnn
  • hartnn
\(\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\)
PhoenixFire
  • PhoenixFire
Okay, I have never been taught such things. I'll give that a go.
PhoenixFire
  • PhoenixFire
For \(ln(t)\gt 0\) \[\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\]\[=ln(2)-\lim \limits_{y \rightarrow 0}ln(y)\] For \(ln(t)\lt 0\) \[\lim \limits_{y \rightarrow 0} \int \limits_y^2-dx/x\]\[=\lim \limits_{y \rightarrow 0}ln(y)-ln(2)\] The limit of \(ln(y)\) as \(y\rightarrow 0\) is \(-\infty\). So how does this work? I have two solutions both at infinity... or have I messed up somewhere?
PhoenixFire
  • PhoenixFire
@hartnn

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