求不定积分xdx/(5+x-x^2)^0.5

Question

求不定积分xdx/(5+x-x^2)^0.5

Answer 1

∫ x/√(5+x-x²) dx
= ∫ x/√[(21/4) - (x-1/2)²] dx
u = x-1/2,du = dx
= ∫ (u+1/2)/√(21/4 - u²) du
u = (1/2)√21 * sinθ,du = (1/2)√21 * cosθ dθ
√(21/4 - u²) = (1/2)√21 * cosθ
原式= ∫ [1/2 + (1/2)√21 * sinθ] dθ
= (1/2)∫ dθ + (1/2)√21 * ∫ sinθ dθ
= (1/2) * θ + (1/2)√21 * (-cosθ) + c
= (1/2) * arcsin(2u / √21) - (1/2)√(21-4u²) + c
= -√(5+x-x²) - (1/2)arcsin[(2x-1) / √21] + c