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AmTran_Bus

  • one year ago

Can someone check this integration?

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  1. AmTran_Bus
    • one year ago
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  2. anonymous
    • one year ago
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    What have you tried? By parts?

  3. AmTran_Bus
    • one year ago
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    Sorry, just got back to this page. Let me show you.

  4. AmTran_Bus
    • one year ago
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    So if you say |dw:1434570936513:dw|

  5. AmTran_Bus
    • one year ago
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    That is after IBP

  6. AmTran_Bus
    • one year ago
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    Then can't you do a sub?

  7. anonymous
    • one year ago
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    So it looks like you set \(u=\ln x\) and \(dv=\dfrac{x}{\sqrt{x^2-1}}\,dx\). A trig substitution might work. Try \(x=\sec u\).

  8. AmTran_Bus
    • one year ago
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    That give|dw:1434571156177:dw|

  9. anonymous
    • one year ago
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    Let's see... \[\int\frac{\sqrt{x^2-1}}{x}\,dx=\int\frac{\sqrt{\sec^2u-1}}{\sec u}\sec u\tan u\,du=\int \tan^2u\,du\] Now a trig identity will work well here: \[\tan^2u=\sec^2u-1\] and you know the antiderivative of \(\sec^2u\).

  10. AmTran_Bus
    • one year ago
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    Yes, so does that not give

  11. AmTran_Bus
    • one year ago
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    |dw:1434571377468:dw|

  12. anonymous
    • one year ago
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    That's right.

  13. AmTran_Bus
    • one year ago
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    Moving on then, I end up at the end getting something different :(|dw:1434571498405:dw|

  14. anonymous
    • one year ago
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    \(\ln x(x-1)\) should be \((\ln x-1)\).

  15. AmTran_Bus
    • one year ago
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    Whoops! Ok. But does that match any of those possible answers, or have we made a mistake?

  16. anonymous
    • one year ago
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    Other than that, your solution is right. I suspect a typo in the answer choices.

  17. AmTran_Bus
    • one year ago
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    Thanks. I wonder if any are equivalent?

  18. AmTran_Bus
    • one year ago
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    Because I have not ever had a problem where they have made a mistake before.

  19. anonymous
    • one year ago
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    You can differentiate the answer choices to check. The second one can be eliminated right away since it doesn't generate any log terms.

  20. AmTran_Bus
    • one year ago
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    Ok. Thanks so much!

  21. anonymous
    • one year ago
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    You're welcome!

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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