Integral of x/(x^4+x^2+1): Solution Attempts

[mwaso]

Homework Statement

the indefinite integral of x/(x^4+x^2+1)

Homework Equations

n/a

The Attempt at a Solution

I didn't see an obvious u-substitution and it didn't look like a partial fractions candidate to me since the bottom is not easily factored. It doesn't look like any of the inverse trig forms. I didn't see how it could be integrated by parts, so I was pretty much lost.

I tried letting u = x^2 so that du would = 2xdx in order to get rid of the x in the numerator, but that didn't really help me as I was left with 1/(u^2+u+1) and that isn't really much better than the original problem

I also tried completing the square in the denominator to get (x^2+1)^2-x^2 my idea was to then allow u to =(x^2+1) and du would equal 2xdx. this left me with the exact same integral as before (of course) and didn't help.

I'm stuck. I'm almost sure that there has to be a substitution I can use to get the integral, but I haven't the faintest idea what it is. I can't see anything that looks promising.

 

[neutrino]

I tried letting u = x^2 so that du would = 2xdx in order to get rid of the x in the numerator, but that didn't really help me as I was left with 1/(u^2+u+1) and that isn't really much better than the original problem

Really? Why not try completing the square at this stage?

 

[mwaso]

because if I completed the square at this stage, I'd get the indefinite integral of du/(u+1)^2-u, right? If I do that, I don't see this would help as it still isn't an inverse trig form and I still don't see any candidates for another substitution...

 

[neutrino]

because if I completed the square at this stage, I'd get the indefinite integral of du/(u+1)^2-u, right?

$$u^2+u+1$$ can also be written as $$\left(u+\frac{1}{2}\right)^2 + \frac{3}{4}$$

In such cases, the "trick" in completing the square, given a quadratic polynomial, is to divide the coefficient of the middle term (when it is written in the usual form ax^2+bx+c) by 2 and then subtracting (b/2)^2 from the constant c.

 

[mwaso]

ahh, wow, that works MUCH better. so that basically leaves me with an inverse tangent, right? good deal. Thanks