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henriquerb

  • 2 years ago

Integrate dx/sqrt(5-2x+x²)

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  1. Eleven17
    • 2 years ago
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    \[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5 - 2x + x^{2}} }\]

  2. henriquerb
    • 2 years ago
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    Yes, thats what I want lol.

  3. Eleven17
    • 2 years ago
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    Haha, Okay! I was going to be rude and just post that picture of it, but I'll actually solve it out for you, do you want any of the work shown?

  4. henriquerb
    • 2 years ago
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    Hahaha, yes, I can't solve that one. You knw when your brain is blocked? If you could do that one I would apreciate.

  5. Eleven17
    • 2 years ago
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    No problem! I love doing Calculus...such a nerd.

  6. Eleven17
    • 2 years ago
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    Do you know integration by parts?

  7. henriquerb
    • 2 years ago
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    Yes, I do.

  8. EulerGroupie
    • 2 years ago
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    I came up with two methods. Both started with completing the square under the radical. After that you can either do a trig. sub. (the long way), or do a u sub. resulting in an inverse hyperbolic trig function (the short way).

  9. henriquerb
    • 2 years ago
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    EulerGroupie, x²-2x+5=(x-1)²+4, is that what you mean by completing squares? (I believe so). I can complete squares, but I'm always confused with that 4 in the sub's. Could you help me?

  10. EulerGroupie
    • 2 years ago
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    Yes, you completed the square properly.\[\int\limits_{}^{}\frac{1}{\sqrt{(x-1)^{2}+4}}dx\]The first step is to recognize that we have an integral that looks like the generalized formula...\[\sinh ^{-1}u+C=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du\]Then we use algebra to get it closer to this form by factoring a 4 from the radicand and eventually taking it out of the radical as its root...\[\int\limits_{}^{}\frac{1}{\sqrt{4[\frac{(x-1)^{2}}{4}+1]}}dx=\int\limits_{}^{}\frac{1}{2\sqrt{(\frac{x-1}{2})^{2}+1}}dx\]Notice how the four left inside is incorporated into the squared term. Now we can use a u-sub where...\[u=\frac{x-1}{2};du=\frac{1}{2}dx;2du=dx\]\[\int\limits_{}^{}\frac{2du}{2\sqrt{u ^{2}+1}}=\int\limits_{}^{}\frac{1}{\sqrt{u ^{2}+1}}du=\sinh ^{-1}u+C\]\[=\sinh ^{-1}(\frac{x-1}{2})+C\]

  11. Krishnadas
    • 2 years ago
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    |dw:1355917276626:dw|

  12. Krishnadas
    • 2 years ago
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    now you can do..

  13. Krishnadas
    • 2 years ago
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    @EulerGroupie isnt \[\int\limits_{}^{} dx/\sqrt{x ^{2}+a ^{2}}\]

  14. Krishnadas
    • 2 years ago
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    @EulerGroupie is \[\log_{}|x+\sqrt{x ^{2}+a ^{2}} ?\]

  15. EulerGroupie
    • 2 years ago
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    \[\ln(x+\sqrt{x ^{2}+1})=\sinh ^{-1}(x)\]I came across the log solution as well, but when I tried to double check my answer with Wolfram, it wasn't there. It offered up the inverse hyperbolic sine solution so I reviewed that section in my Calc text and sure enough, they are equivalent. I thought the the sinh seemed more elegant (and verifiable through Wolfram), so I went with it.

  16. Krishnadas
    • 2 years ago
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    @EulerGroupie when the lesson on integration was taken on my class,we studied only that equation...thats why...there must be many answers to one question in integration...we both are correct i think..

  17. EulerGroupie
    • one year ago
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    absolutely

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