(1)：反正切里面那个应该是√x而不是x,否则凑不到微分的.
∫ arctan√x/[(1 + x)√x] dx
= 2∫ arctan√x/[(1 + x) * 2√x] dx
= 2∫ arctan√x/(1 + (√x)²) d√x
= 2∫ arctan√x d(arctan√x)
= (arctan√x)² + C
(2)：令u = (1 + x)^(1/3),u³ = 1 + x,3u² du = dx
∫ 1/[1 + (1 + x)^(1/3)] dx,x开三次方 = x^(1/3)
= ∫ 1/(1 + u) * (3u² du)
= 3∫ u[(u + 1) - 1]/(1 + u) du
= 3∫ u du - 3∫ u/(1 + u) du
= (3/2)u² - 3∫ [(u + 1) - 1]/(1 + u) du
= (3/2)u² - 3∫ du + 3∫ du/(1 + u)
= (3/2)u² - 3u + 3ln|1 + u| + C
= (3/2)(1 + x)^(2/3) - 3(1 + x)^(1/3) + 3ln|1 + (1 + x)^(1/3)| + C
(3)：这题可说能秒杀
∫ x²/(x³ + 3) dx
= ∫ 1/(x³ + 3) d(x³/3)
= (1/3)∫ 1/(x³ + 3) d(x³ + 3)
= (1/3)ln|x³ + 3| + C