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

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \int \frac{\mathrm dx}{x\sqrt{x^2+x+1}}\]

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

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

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

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

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

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

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2it'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
 one year ago
Best ResponseYou've already chosen the best response.0this might work ... http://en.wikipedia.org/wiki/Euler_substitution this is uglier than weirstrass substitution.

sirm3d
 one year ago
Best ResponseYou've already chosen the best response.2i 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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