## sirm3d 2 years ago integral problem. i have seen one solution in full. Maybe i can get another solution here.

1. Spacelimbus

$\Large \int \frac{x}{\sqrt{x^2+x+1}}dx$?

2. sirm3d

$\Large \int \frac{\mathrm dx}{x\sqrt{x^2+x+1}}$

3. Spacelimbus

ah, I see.

4. TuringTest

This integral is frustrating...

5. Spacelimbus

$\Huge \checkmark$

6. Spacelimbus

What did you try @TuringTest, completing the square and then a trig substitution?

7. Spacelimbus

That's what I did, and I did end up with an integral that is optically neater, integration wise still as bad

8. 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

9. TuringTest

From there I tried some difference of squares voodoo, but that only serves to complicate things it seems :p

10. 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.

11. experimentX

this might work ... http://en.wikipedia.org/wiki/Euler_substitution this is uglier than weirstrass substitution.

12. 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 $y=\frac{1}{2} + \frac{1}{x}\\\mathrm dy=-\frac{1}{x^2}\mathrm dx$