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sirm3d

  • 3 years ago

integral problem. i have seen one solution in full. Maybe i can get another solution here.

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  1. Spacelimbus
    • 3 years ago
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    \[\Large \int \frac{x}{\sqrt{x^2+x+1}}dx  \]?

  2. sirm3d
    • 3 years ago
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    \[\Large \int \frac{\mathrm dx}{x\sqrt{x^2+x+1}}\]

  3. Spacelimbus
    • 3 years ago
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    ah, I see.

  4. TuringTest
    • 3 years ago
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    This integral is frustrating...

  5. Spacelimbus
    • 3 years ago
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    \[\Huge \checkmark \]

  6. Spacelimbus
    • 3 years ago
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    What did you try @TuringTest, completing the square and then a trig substitution?

  7. Spacelimbus
    • 3 years ago
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    That's what I did, and I did end up with an integral that is optically neater, integration wise still as bad

  8. TuringTest
    • 3 years ago
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    yep, exactly. then I wound up with\[2\int\frac{\sec\theta d\theta}{\sqrt3\tan\theta-1}\]at which point I'm stuck

  9. TuringTest
    • 3 years ago
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    From there I tried some difference of squares voodoo, but that only serves to complicate things it seems :p

  10. sirm3d
    • 3 years ago
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    it's the same solution i had. \[2\frac{\sec \theta}{\sqrt 3 \tan \theta -1}=\frac{1}{\tfrac{\sqrt 3}{2} \sin \theta -\tfrac{1}{2}\cos \theta}=\frac{1}{\sin(\theta-\pi/6)}\] which can be integrated easily but the return to the variable \(x\) is the frustrating part.

  11. experimentX
    • 3 years ago
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    this might work ... http://en.wikipedia.org/wiki/Euler_substitution this is uglier than weirstrass substitution.

  12. sirm3d
    • 3 years ago
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    i found another solution. \[\frac{1}{x\sqrt{x^2+x+1}}=\frac{1}{\displaystyle x^2\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}}=\frac{1}{\displaystyle x^2\sqrt{\left(\frac{\sqrt 3}{2}\right)^2+\left(\frac{1}{2}+\frac{1}{x}\right)^2}}\] let \[y=\frac{1}{2} + \frac{1}{x}\\\mathrm dy=-\frac{1}{x^2}\mathrm dx\]

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