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PhoenixFire

  • 2 years ago

\[\int_{e^-1}^{e} {\frac{1}{t(1+\left | ln(t) \right |)} dt}\] 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.

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  1. divu.mkr
    • 2 years ago
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    tell me what have you done untill now?

  2. hartnn
    • 2 years ago
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    substitute 1+ln t = x hace u donee this ?

  3. hartnn
    • 2 years ago
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    **have done

  4. hartnn
    • 2 years ago
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    and if u have done that, have you changed the limits ?

  5. PhoenixFire
    • 2 years ago
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    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.

  6. divu.mkr
    • 2 years ago
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    then take the limits of 0... that what we do..

  7. hartnn
    • 2 years ago
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    \(\lim \limits_{y \rightarrow 0} \int \limits_y^2dx/x\)

  8. PhoenixFire
    • 2 years ago
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    Okay, I have never been taught such things. I'll give that a go.

  9. PhoenixFire
    • 2 years ago
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    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?

  10. PhoenixFire
    • 2 years ago
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    @hartnn

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