不定积分∫x^2⼀（1+x^2）^2dx

∫ x²/(1 + x²)² dx,令x = tanz,dx = sec²z dz
= ∫ tan²z/sec⁴z * (sec²z dz)
= ∫ sin²z/cos²z * cos²z dz
= ∫ (1 - cos2z)/2 dz
= z/2 - (1/4)sin2z + C
= (1/2)arctanx - (1/2) * x/√(1 + x²) * 1/√(1 + x²) + C
= (1/2)arctanx - x/[2(1 + x²)] + C

1+x-x^2 = 5/4-(x-1/2)^2
let
x-1/2 = (√5/2)sinu
dx =(√5/2)cosu du
∫x/√(1+x-x^2)dx
=-(1/2)∫(1-2x)/√(1+x-x^2)dx +(1/2)∫dx/√(1+x-x^2)
=-(1/2)√(1+x-x^2) +(1/2)∫dx/√(1+x-x^2)
=-(1/2)√(1+x-x^2) +(√5/5)∫ du
=-(1/2)√(1+x-x^2) +(√5/5)u + C
=-(1/2)√(1+x-x^2) +(√5/5)arcsin[(2x-1)/√5] +C