Question

如何求 x2/根号下(1 x-x2)的不定积分

如何求 x2/根号下(1 +x-x2)的不定积分

Answer 1

∫ x²/√(1 + x - x²) dx
= ∫ x²/√[5/4 - (x - 1/2)²] dx
令x - 1/2 = ρ，a² = 5/4
= ∫ (ρ + 1/2)²/√(a² - ρ²) dρ
令ρ = asinz，dρ = acosz dz
= ∫ (asinz + 1/2)²/(acosz) * (acosz dz)
= ∫ (a²sin²z + asinz + 1/4) dz
= (a²/2)∫ (1 - cos2z) dz + a∫ sinz dz + (1/4)∫ dz
= (a²/2)[z - sinzcosz] - acosz + z/4 + C
= (a²/2 + 1/4)z - (a²/2)sinzcosz - acosz + C
= (1/4)(2a² + 1)arcsin(ρ/a) - (a²/2)(ρ/a)√(a² - ρ²)/a - a√(a² - ρ²)/a + C
= (- 1/4)(2x + 3)√(1 + x - x²) - (7/8)arcsin[(1 - 2x)/√5] + C