Question

定积分∫﹙1,4﹚1/﹙1＋√x﹚dx=？

Answer 1

用凑微分法：
∫(1到4) 1 / √(1 + √x) dx
= ∫ 2√x / [2√x * (1 + √x)] dx，上下乘以2√x，因为1/(2√x) = (√x)'
= 2∫ √x / (1 + √x) d√x
= 2∫ (√x + 1 - 1) / (1 + √x) d√x
= 2∫ [1 - 1 / √(1 + √x)] d√x
= 2∫ d√x - 2∫ d(√x + 1) / (1 + √x)
= 2√x - 2ln|1 + √x|
= [2√4 - 2ln(1+√4)] - [2√1 - 2ln(1+√1)]
= 4 - 2ln3 - 2 + 2ln2
= 2 + 2ln2 - 2ln3
= 2[1 + ln(2/3)]

用换元法：
设u = √x，x = u²，dx = 2u du
当x = 1，u = √1 = 1
当x = 4，u = √4 = 2
所以新的积分区间变为1到2
∫ 2u / (1 + u) * du
= 2∫ (u + 1 - 1) / (1 + u) du
= 2∫ [1 - 1 / (1 + u)] du
= 2∫ du - 2∫ d(u + 1) / (1 + u)
= 2u - 2ln|1 + u|
= [2*2 - 2ln(1+2)] - [2*1 - 2ln(1+1)]
= 4 - 2ln3 - 2 + 2ln2
= 2 + 2ln2 - 2ln3
= 2[1 + ln(2/3)]
或= 2ln(2e/3)

Answer 2

换元√x=t t由1到2
=∫(1,2) 2t/(1+t) dt
=∫(1,2) (2t+2-2)/(1+t) dt
=(2t+2ln(1+t)) (1,2)
=2-2ln(3/2)

Answer 3

∫﹙1,4﹚1/﹙1＋√x﹚dx
设√x=a，则
原始=∫﹙1,2﹚1/﹙1＋a﹚da^2
=∫﹙1,2﹚2/﹙1＋a﹚da
=2∫﹙1,2﹚1/﹙1＋a﹚d(1+a)
=2∫﹙1,2﹚dln(a+1)
=2*(ln3-ln2)