求定积分(1~0)√x/1+√xdx详细
问题描述：
求定积分(1~0)√x/1+√xdx详细
答
令a=√x
x=a²
dx=2ada
所以原式=∫(0,1)a/(1+a)*2ada
=2∫(0,1)(a²-1+1)/(a+1) da
=2∫(0,1)[a-1+1/(a+1)]da
=2[a²/2-a+ln|1+a|] (0,1)
=2[(1/2-1+ln2)-(0-0+0)]
=-1+2ln2
答
I = ∫√xdx/(1+√x)
= ∫2xd√x/(1+√x)
= 2∫（x-1+1）d√x/(1+√x)
= 2∫（√x-1）d√x + 2∫d(1+√x)/(1+√x)
= 2[x/2-√x] + 2[ln(1+√x]
= -1+2ln2