x^2/(1x-x^2)的不定积分.擦.怎么搞得.x^2/√(1+x-x^2)

Question

x^2/(1 x-x^2)的不定积分.
擦.怎么搞得.x^2/√(1+x-x^2)

Answer 1

K = ∫ x²/√(1+x-x²) dx
= ∫ x²/√[5/4-(x-1/2)²] dx
Let x-1/2 = √5/2*sinθ then dx = √5/2*cosθdθ
√[5/4-(x-1/2)²] = √(5/4-5/4*sin²θ) = √[(5/4)(cos²θ)] = √5/2 * cosθ
x² = (√5/2*sinθ+1/2)² = (5/4)sin²θ+(√5/2)sinθ+1/4
K = ∫ [(5/4)sin²θ+(√5/2)sinθ+1/4]/(√5/2 * cosθ) * (√5/2 * cosθ dθ)
= (5/4)∫ sin²θ dθ + (√5/2)∫ sinθ dθ + (1/4)∫ dθ
= (5/8)∫ (1-cos2θ) dθ + (√5/2)(-cosθ) + (1/4)θ
= (5/8)(θ-1/2*sin2θ) - (√5/2)(cosθ) + (1/4)θ + C
x-1/2 = √5/2*sinθ,sinθ = (2x-1)/√5,cosθ = 2√(1+x-x²)/√5
K = (5/8+1/4)arcsin[(2x-1)/√5] - (5/8) * (2x-1)/√5 * 2√(1+x-x²)/√5 - (√5/2) * 2√(1+x-x²)/√5 + C
= -(1/4)(2x-1)√(1+x-x²) - √(1+x-x²) + (7/8)arcsin[(2x-1)/√5)] + C