Question

求定积分(1~0)√x/1+√xdx详细

Answer 1

I = ∫<下限0,上限1>√xdx/(1+√x)
= ∫<0,1>2xd√x/(1+√x)
= 2∫<0,1>（x-1+1）d√x/(1+√x)
= 2∫<0,1>（√x-1）d√x + 2∫<0,1>d(1+√x)/(1+√x)
= 2[x/2-√x]<0,1> + 2[ln(1+√x]<0,1>
= -1+2ln2

Answer 2

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