Integration help?

Question

I need to find the integral of:
x / (x^4 + x^2 + 1) dx

I was thinking possibly partial fractions, but I'm not quite sure how to factor the denominator. Any suggestions? Thanks.

Answer 1

I = ∫x / (x^4 + x^2 + 1) dx

Ordinarily yes, but here you can't do partial fractions, if the denominator has no factors.
(You probably were thinking that (x^4 + x^2 + 1) is itself a FACTOR in (x^6-1)/(x^2-1) = (x^4 + x^2 + 1).)

[Ok the guy below did find factors for the denominator, so I was wrong. Using what he supplied:
x^4 + x² + 1 = (x²+x+1)(x²-x+1)

Then x/x^4 + x² + 1 = (Ax+B)/(x²+x+1) + (Cx+D)/(x²-x+1)
x = (Ax+B)*(x²-x+1) + (Cx+D)*(x²+x+1)
= A(x³-x²+x) +B(x²-x+1) + C(x³+x²+x) + D(x²+x+1)

Taking coeffts:
[x³] A+C=0 => C= -A
[1] B+D=0 => D= -B
[x²] -A+B+C+D=0 =>-2A=0 => A=0,C=0
[x] A-B+C+D=1 => -2B=1=>B= -1/2, D = +1/2

Then x/x^4 + x² + 1 = 1/2(x²-x+1) - 1/2(x²+x+1)

I = ∫x / (x^4 + x^2 + 1) dx

I = 1/2∫1/(x²-x+1) - 1/(x²+x+1) dx
completing the square in each denominator...
I = 1/2∫1/((x-1/2)² +5/4) - 1/((x+1/2)² +3/4) dx
I = 1/2∫dx/((x-1/2)² +(√5/2)²) - dx/((x+1/2)² +(√3/2)²)

For a denominator x²+a² you still need the trigonometric substitutions
x = a tan θ => I = 1/a arctan(x/a)
or in these two cases:
(x-1/2)= (√5/2) tan θ => I1 = (2/√5) arctan (x-1/2 / √5/2)
(x+1/2)= (√3/2) tan θ=> I2 = (2/√3) arctan (x-1/2 / √3/2)

Finally
I = (1/√5) arctan (2x-/√5) - (1/√3) arctan (2x-1/ √3) + C

_____________________________
Earlier work:
I = ∫x / (x^4 + x^2 + 1) dx
so we can transform to
I = ∫x (x^2-1)/(x^6-1) dx
**IFF the domain of x does not include the singularity x=±1 **
but this doesn't get us anywhere much:
I = ∫x (x^2-1)/(x^6-1) dx
I = ∫(x²-1)/[(x²-1)³ +3x^4 -3x²] (xdx)
I = ∫(x²-1)/[(x²-1)³ +3x²(x²-1)] (xdx)
I = ∫1/[(x²-1)² +3x²] (xdx)

So back to the original:
I = ∫x / (x^4 + x^2 + 1) dx

Now substitute u = x²
(u = x²-1 seemed promising but wasn't really)
du/2 = xdx
I = ∫u/(u³+3(u+1)u) (du/2)
I = ∫1/(u²+3(u+1)) (du/2)


So it's a plain old two-step power-and-trigonometric substitution:
I = ∫x / (x^4 + x^2 + 1) dx
Let u=x², du = 2xdx
I = ∫(du/2)/(u² + u + 1)

Answer 2

To get you started, write
x^4 + x² + 1 = x^4 + 2x² + 1 - x² =
(x²+1)²-x² = (x²+x+1)(x²-x+1).
Now use partial fractions, as you suggested.