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Lammy Group Title

find integrate (1/(xsqrt(1+(lnx)^2)),1,e,x)

  • 2 years ago
  • 2 years ago

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  1. Lammy Group Title
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    \[\int\limits_{1}^{e}1/(x \sqrt{1+(lnx)^2}) dx\]

    • 2 years ago
  2. myininaya Group Title
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    \[\int\limits_{1}^{e}\frac{dx}{x \sqrt{1+(\ln(x))^2}}\] \[\text{ Let } a=\ln(x) => da=\frac{dx}{x}\] if x=e then a=ln(e)=1 if x=1 then a=ln(1)=0 So we have \[\int\limits_{0}^{1}\frac{da}{\sqrt{1+a^2}} \]

    • 2 years ago
  3. shaan_iitk Group Title
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    ln x= t .. then this becomes dt/(sqrt(1+t^2)) .. now put t = tany and you get secydy = ln(secy + tany)

    • 2 years ago
  4. myininaya Group Title
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    \[\text{ Let } a=\tan(\theta) => da=\sec^2(\theta) d \theta\] if a=1 then theta=arctan(1)=pi/4 if a=0 then theta=arctan(0)=0 \[\int\limits_{0}^{\frac{\pi}{4}}\frac{\sec^2(\theta) d \theta}{\sqrt{1+\tan^2(\theta)}}\]

    • 2 years ago
  5. Lammy Group Title
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    yea thats as far as i got and lost

    • 2 years ago
  6. myininaya Group Title
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    \[\int\limits_{0}^{\frac{\pi}{4}}\sec(\theta) d \theta \text{ (note :} 1+\tan^2(\theta) =\sec^2(\theta) \text{ )}\]

    • 2 years ago
  7. myininaya Group Title
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    \[=\ln|\sec(\theta)+\tan(\theta)||_0^\frac{\pi}{4}=\ln|\sec(\frac{\pi}{4})+\tan(\frac{\pi}{4})|-\ln|\sec(0)+\tan(0)|\]

    • 2 years ago
  8. myininaya Group Title
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    \[=\ln(\frac{1}{\frac{\sqrt{2}}{2}}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}})-\ln(\frac{1}{1}+\frac{0}{1})\]

    • 2 years ago
  9. myininaya Group Title
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    \[=\ln(\frac{2}{\sqrt{2}}+1)-\ln(1)=\ln(\frac{2 \sqrt{2}}{2}+1)-0=\ln(\sqrt{2}+1)\]

    • 2 years ago
  10. Lammy Group Title
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    for the absolute values, how do you get rid of it?

    • 2 years ago
  11. Lammy Group Title
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    thanks for your help! very clear instruction. i always liked your help =)

    • 2 years ago
  12. myininaya Group Title
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    the number inside was positive you don't need them anymore

    • 2 years ago
  13. myininaya Group Title
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    \[\ln|1|=\ln(1)\] You don't need the absolute value since |1|=1

    • 2 years ago
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