I = ∫ (1/(x^5 +1)) dx = ∫ (1/((x+1)(x^4 −x^3 +x^2 −x+1))) dx = ∫ [(1/(5(x+1))) − ((x^3 −2x^2 +3x−4)/(5(x^4 −x^3 +x^2 −x+1)))] dx = (1/5)∫ (1/(x+1)) dx − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = (1/5)ln ∣x+1∣ + C_1 − (1/5)∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx I_2 = ∫ ((x^3 −2x^2 +3x−4)/(x^4 −x^3 +x^2 −x+1)) dx = ∫ ((x^3 −2x^2 +3x−4)/((x^2 +(ϕ−1)x+1)(x^2 −ϕx+1))) dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1)) + ((ϕx−2)/(x^2 −ϕx+1))] dx = ∫ [(((1−ϕ)x−2)/(x^2 +(ϕ−1)x+1))] dx + ∫ [((ϕx−2)/(x^2 −ϕx+1))] dx = ((1−ϕ)/2)ln ∣x^2 +(ϕ−1)x+1∣ − (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) − (ϕ/2)ln ∣x^2 −ϕx+1∣ − (√(3−ϕ))tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C_2 I = determinant ((((1/5)ln ∣x+1∣ − ((1−ϕ)/(10))ln ∣x^2 +(ϕ−1)x+1∣ + (√(2−ϕ))tan^(−1) (((2x+ϕ−1)/( (√(2+ϕ))))) + (ϕ/(10))ln ∣x^2 −ϕx+1∣ + ((√(3−ϕ))/5)tan^(−1) (((2x−ϕ)/( (√(3−ϕ))))) + C))) Where ϕ = Golden Ratio = (((√5) + 1)/2) I wonder if I made any mistakes here. Can someone check this?