求定积分(1~0)√x/1+√xdx详细

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求定积分(1~0)√x/1+√xdx详细
求定积分(1~0)√x/1+√xdx详细

求定积分(1~0)√x/1+√xdx详细
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

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