Integral of 1+x2/(1+x4): Step-by-Step Solution

[mjolnir80]

Homework Statement

find the indefinate integral of 1+x²/(1+x⁴)

Homework Equations

The Attempt at a Solution

im having trouble trying to figure what to use to substitute into this equation

 

[Dick]

I would do it by partial fractions. You can factor (x^4+1) into two real quadratics. The coefficients may not all be rational, but that's ok. Can you do that?

 

[Mark44]

I would do it by partial fractions. You can factor (x^4+1) into two real quadratics. The coefficients may not all be rational, but that's ok. Can you do that?

I'd like to see that factorization. I've been under the impression for a long time that x^4 + 1 was irreducible over the reals.

If we allow complex numbers, the fourth roots of -1 (solutions of x^4 = -1) are
cos(pi/4) + i sin(pi/4)
cos(3pi/4) + i sin(3pi/4)
cos(5pi/4) + i sin(5pi/4)
cos(7pi/4) + i sin(7pi/4)
none of which are real.

 

[weejee]

(1+x^2)/(1+x^4) = 1/(1+(sqrt(2)*x+1)^2) + 1/(1+(sqrt(2)*x-1)^2)

I found this by tracking back from the answer, which is obtained using Maple.

For how to get this relation without referring to the answer, I'm not sure.

 

[Dick]

I'd like to see that factorization. I've been under the impression for a long time that x^4 + 1 was irreducible over the reals.

If we allow complex numbers, the fourth roots of -1 (solutions of x^4 = -1) are
cos(pi/4) + i sin(pi/4)
cos(3pi/4) + i sin(3pi/4)
cos(5pi/4) + i sin(5pi/4)
cos(7pi/4) + i sin(7pi/4)
none of which are real.

Ok. x^4+1=(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1). x^4+1 is irreducible over the RATIONALS, not over the REALS. Your roots group into two pairs of complex conjugates (of course). Use them to construct the quadratics.

 

[Mark44]

Ok. x^4+1=(x^2+sqrt(2)x+1)(x^2-sqrt(2)+1). x^4+1 is irreducible over the RATIONALS, not over the REALS. Your roots group into two pairs of complex conjugates (of course). Use them to construct the quadratics.

OK, let me rephrase: x^4 + 1 can't be factored into linear factors with real coefficients.