Question

求不定积分∫√((2-x)/(x+1))dx

Answer 1

√[(2-x) / (x+1)] = (2-x) / √[(2-x)(x+1)] = (2-x) / √(-x^2 + x + 2) = (2-x) / √[-(x - 1/2)^2 + 9/4]
= 2/3 * (2-x) / √[1 - 4/9 * (x - 1/2)^2]
令 2/3 * (x - 1/2) = sin t => x = 3/2 * sin t + 1/2 => dx = 3/2 * cos t dt, t = arcsin[2/3 * (x - 1/2) ]
= arcsin[2x/3 - 1/3], cos t = √(1 - (sin t)^2) = 2/3 * √(-x^2 + x + 2).
原积分 = ∫ 2/3 * (2 - 3/2 * sin t - 1/2) / cos t * 3/2 * cos t dt
= 3/2 * ∫ (1 - sin t) dt
= 3/2 * (t + cos t) + C,C为任意常数.
= 3/2 * (arcsin[2x/3 - 1/3] + 2/3 * √(-x^2 + x + 2) + C
= 3/2 * arcsin[2x/3 - 1/3] + √(-x^2 + x + 2) + C.