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sirm3d
 3 years ago
integral problem. i have seen one solution in full. Maybe i can get another solution here.
sirm3d
 3 years ago
integral problem. i have seen one solution in full. Maybe i can get another solution here.

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

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

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

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

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

Spacelimbus
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years 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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