Question

∫ sin√xdx 分步积分 ∫(1+x)dx/(1+√x) 换元法

Answer 1

∫ sin√x · (2√x)/(2√x) dx
= 2∫ √x sin√x d(√x)
= - 2∫ √x d(cos√x)
= - 2√x cos√x + 2∫ cos√x d√x <==分部积分法
= - 2√x cos√x + 2sin√x + C
∫ (1 + x)/(1 + √x) dx
令u = √x,x = u²,dx = 2u du
= ∫ (1 + u²)/(1 + u) · 2u du
= 2∫ u(1 + u²)/(1 + u) du
= 2∫ [u² - u + 2 - 2/(u + 1)] du <==综合除法
= 2 · [u³/3 - u²/2 + 2u - 2ln|u + 1|] + C
= (2/3)x^(3/2) - x + 4√x - 4ln|1 + √x| + C <==将u = √x回代