sirm3d
  • sirm3d
integral problem. i have seen one solution in full. Maybe i can get another solution here.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\Large \int \frac{x}{\sqrt{x^2+x+1}}dx  \]?
sirm3d
  • sirm3d
\[\Large \int \frac{\mathrm dx}{x\sqrt{x^2+x+1}}\]
anonymous
  • anonymous
ah, I see.

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TuringTest
  • TuringTest
This integral is frustrating...
anonymous
  • anonymous
\[\Huge \checkmark \]
anonymous
  • anonymous
What did you try @TuringTest, completing the square and then a trig substitution?
anonymous
  • anonymous
That's what I did, and I did end up with an integral that is optically neater, integration wise still as bad
TuringTest
  • TuringTest
yep, exactly. then I wound up with\[2\int\frac{\sec\theta d\theta}{\sqrt3\tan\theta-1}\]at which point I'm stuck
TuringTest
  • TuringTest
From there I tried some difference of squares voodoo, but that only serves to complicate things it seems :p
sirm3d
  • sirm3d
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.
experimentX
  • experimentX
this might work ... http://en.wikipedia.org/wiki/Euler_substitution this is uglier than weirstrass substitution.
sirm3d
  • sirm3d
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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