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Math Help - Please help integrate...

  1. #1
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    Please help integrate...

    I need some help with integrating {1 dx / (x^2+3)^(3/2) }. I got
    (1/3)[ln {(x^2+3)^(1/2) +x} - ln {(3)^(1/2)}] + C. I tried to differentiate this but could not get it back to original form.
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  2. #2
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    \int\frac1{(x^2+3)^{3/2}}~dx

    Did you try the substitution x=\sqrt3\tan{u} ?
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  3. #3
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    Quote Originally Posted by Possible actuary View Post
    I need some help with integrating {1 dx / (x^2+3)^(3/2) }. I got
    (1/3)[ln {(x^2+3)^(1/2) +x} - ln {(3)^(1/2)}] + C. I tried to differentiate this but could not get it back to original form.
    \int \frac{1}{(x^2+3)^{3/2}} \ dx

    I am going to do this in a strange way:
    Instead find,
    \int \frac{1}{(x^2+1)^{3/2}} dx = \int \frac{1}{\sqrt{x^2+1}}\cdot \frac{1}{x^2+1} \ dx

    Use substitution, t = \tan^{-1} x \Rightarrow t' = \frac{1}{x^2+1}
    With \sec t = \frac{1}{\sqrt{x^2+1}}

    Thus,
    \int \sec t dt = \ln |\sec t + \tan t| + C

    Thus,
    \ln | \sec (\tan^{-1} x) + \tan (\tan^{-1} x) | + C = \ln | \sqrt{x^2+1} + x | + C

    Apparently this can also be done with inverse hyperbolic functions, but I think it is a waste to memorize them. So I did it the standard way.

    Now given,
    \int \frac{1}{(x^2+3)^{3/2}} dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left( \frac{x^2}{3} + 1 \right)^{3/2}} \ dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left[ \left( \frac{x}{\sqrt{3}}\right)^2 + 1 \right]^{3/2}} \ dx

    Use the substitution u= \frac{x}{\sqrt{3}} and above work to get:

    \frac{1}{3^{3/2}} \cdot \frac{\sqrt{3}}{1} \cdot \ln \left| \sqrt{ \frac{x^2}{3} + 1} + \frac{x}{\sqrt{3}} \right| + C = \frac{1}{3} \cdot \ln \left| \frac{\sqrt{x^2+3}}{\sqrt{3}} + \frac{x}{\sqrt{3}} \right|+C

    Use the properties of the natural logarithm,
     \frac{1}{3} \cdot \left( \ln | \sqrt{x^2+3} + x | - \ln \frac{1}{\sqrt{3}} \right)+C

    Combine constants,
    \frac{1}{3}\cdot \ln |\sqrt{x^2+3} + x |  + C
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  4. #4
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    Yes I did.

    [PHP][/PHP]
    Attached Files Attached Files
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    Set x=\sqrt3\tan{u}\implies{dx=\sqrt3\sec^2u~du}

    The integral becomes to

    \int\frac1{(x^2+3)^{3/2}}~dx=\frac13\int\cos{u}~du=\frac13\sin{u}+k=\fra  c{x}{3\sqrt{x^2+3}}+k,~~k\in\mathbb{R}
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  6. #6
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    I think that the use of logarithmic properties confused me. I know that ln (a/b) = ln a - ln b. I am not for sure on ln a - ln (1/b).

    That post disappeared. Oh, well.
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  7. #7
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    I dunno why The Perfect Hacker deleted his post

    Anyway, did you understand my solution?
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  8. #8
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    Quote Originally Posted by Krizalid View Post
    I dunno why The Perfect Hacker deleted his post

    Anyway, did you understand my solution?
    I can restore it.

    But when I was posting I was doing it without much concentration. So it is possible I made a big mistake. Since your answer does not agree with mine I assumed I made a mistake.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    I can restore it.

    But when I was posting I was doing it without much concentration. So it is possible I made a big mistake. Since your answer does not agree with mine I assumed I made a mistake.
    I think that we had the same answer up till (1/3)[ ln {(x^2+3)^(1/2) + x / sqrt 3} + C. Then you had put (1/3)[ ln {(x^2+3)^(1/2) +x} - ln (1/sqrt 3) when I thought that the last part should be ln (sqrt 3).
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  10. #10
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    Quote Originally Posted by Krizalid View Post
    Set x=\sqrt3\tan{u}\implies{dx=\sqrt3\sec^2u~du}

    The integral becomes to


    \int\frac1{(x^2+3)^{3/2}}~dx=\frac13\int\cos{u}~du=\frac13\sin{u}+k=\fra  c{x}{3\sqrt{x^2+3}}+k,~~k\in\mathbb{R}
    I don't know where you pulled k from or how it is figured.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Possible actuary View Post
    I don't know where you pulled k from or how it is figured.
    it's just the arbitrary constant we get from integrating. we usually call it "C"
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  12. #12
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    Quote Originally Posted by Possible actuary View Post
    I think that we had the same answer up till (1/3)[ ln {(x^2+3)^(1/2) + x / sqrt 3} + C. Then you had put (1/3)[ ln {(x^2+3)^(1/2) +x} - ln (1/sqrt 3) when I thought that the last part should be ln (sqrt 3).
    Okay I restored it. See if it is correct. Because this Whiskey is making it hard to concentrate on the integral. (Do not drink and derive!)
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  13. #13
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    Quote Originally Posted by Jhevon View Post
    it's just the arbitrary constant we get from integrating. we usually call it "C"
    Okay, I know it as C. Thanks!
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