How to integrate x^4/(1+x^4)?

To integrate x^4/(1+x^4), use partial fractions and substitution.

First, factor the denominator: 1+x^4 = (x^2+1)^2 - 2x^2. Then, use partial fractions to write the integrand as A(x^2+1)/(x^2+1)^2 + B(x^2+1)/(x^2+1)^2 - 2Cx/(x^2+1)^2, where A, B, and C are constants.

Simplify the fractions and solve for A, B, and C by equating the numerators: x^4 = A(x^2+1) + B(x^2+1) - 2Cx. This gives A = 1/2, B = 1/2, and C = 1/4.

Substitute u = x^2+1 and du = 2x dx to get the integral in terms of u: ∫(1/2)/(u-2sqrt(u)+2)du + ∫(1/2)/(u+2sqrt(u)+2)du - ∫(1/4)/(u-2sqrt(u)+2)^2 du.

Use substitution again with v = sqrt(u) - 1 and dv = (1/2)u^(-1/2) du to get the first two integrals in terms of v: ∫(1/4)/(v-1)^2 dv + ∫(1/4)/(v+1)^2 dv.

Integrate each of these using the power rule: (-1/4)/(v-1) - (-1/4)/(v+1) = (-1/2)/(sqrt(u)-1) + (1/2)/(sqrt(u)+1).

For the third integral, use substitution with w = sqrt(u) - 1 and dw = (1/2)u^(-1/2) du: ∫(1/4)/(w^2+1)dw = (1/4)arctan(w) = (1/4)arctan(sqrt(x^2+1)-1).

Substitute back in for u and simplify: (-1/2)/(sqrt(x^2+1)-1) + (1/2)/(sqrt(x^2+1)+1) + (1/4)arctan(sqrt(x^2+1