>>777まず
1/(x^3+1)＝(1/3)*[{1/(x+1)}−{(x−2)/(x^2-x+1)}

(1/3)*∫[0,1]{1/(x+1)}dx＝(1/3)*log2

次に
−(1/3)*∫[0,1]{(x−2)/(x^2-x+1)}dx
＝−(1/3)*∫[0,1][{(x−1/2)−3/2}/(x^2-x+1)]dx
＝−(1/3)*∫[0,1]{(x−1/2)/(x^2-x+1)}dx＋(1/2)*∫[0,1]{1/(x^2-x+1)}dx

(x^2-x+1)'＝2x−1，2x−1＝(x−1/2)/２より
−(1/3)*∫[0,1]{(x−1/2)/(x^2-x+1)}dx
＝−(1/6)*∫[0,1]{(x^2-x+1)'/(x^2-x+1)}dx
＝−(1/6)*[log(x^2-x+1)][0,1]
＝０

(x^2-x+1)＝[{ｘ−(1/2)}^2]＋(3/4)より
(1/2)*∫[0,1]{1/(x^2-x+1)}dx
＝(1/2)*∫[0,1]{1/{(x−1/2)^2＋3/4}]dx
ここで
ｘ＝(1/2)＋{(√3)*(tanθ)/2}とおく。dx/dθ＝√3/{2(cosθ)^2}
ｘ：０→１のときθ：−π/6→π/6
(1/2)*∫[0,1]{1/{(x−1/2)^2＋3/4}]dx
＝(1/2)*(4/3)*{(√3)/2}*∫[−π/6,π/6]1/[{(cosθ)^2}*{(tanθ)^2＋1}]dθ
＝(1/√3)*∫[−π/6,π/6]dθ　(∵(√3)/3＝1/√3，(tanθ)^2＋1＝1/{(cosθ)^2})
＝π/(3√3)

∴∫[0,1]{1/(x^3+1)}dx＝(1/3)*log2＋{π/(3√3)}