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

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SpacelimbusBest ResponseYou've already chosen the best response.0
\[\Large \int \frac{x}{\sqrt{x^2+x+1}}dx \]?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
\[\Large \int \frac{\mathrm dx}{x\sqrt{x^2+x+1}}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
This integral is frustrating...
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
\[\Huge \checkmark \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
What did you try @TuringTest, completing the square and then a trig substitution?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.0
That's what I did, and I did end up with an integral that is optically neater, integration wise still as bad
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
yep, exactly. then I wound up with\[2\int\frac{\sec\theta d\theta}{\sqrt3\tan\theta1}\]at which point I'm stuck
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
From there I tried some difference of squares voodoo, but that only serves to complicate things it seems :p
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
this might work ... http://en.wikipedia.org/wiki/Euler_substitution this is uglier than weirstrass substitution.
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
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 \[y=\frac{1}{2} + \frac{1}{x}\\\mathrm dy=\frac{1}{x^2}\mathrm dx\]
 one year ago
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