Question

∫√x/1+√x dx 上限1下限0

∫√x/1+√x dx 上限1下限0
∫√x/（1+√x） dx 上限1下限0

Answer 1

(定积分的换元积分法）

令t=√x，则x=t^2，dx=2tdt

当x=0，t=0; 当x=1,t=1

∫(下限0,上限1)[√x/(1+√x)]dx

=∫(下限0,上限1)[t/(1+t)] 2tdt

=2∫(下限0,上限1)[t -1 + 1/(1+t)] dt

=2[1/2 t^2 - t + ln(1+t)] |(下限0,上限1)

=2ln2 - 1

Answer 2

不定积分：∫√x/(1+√x)dx
=x-2√x+2ln[√x+1]+C

定积分=(1-2√1+2ln[√1+1]+C)-(0-2√0+2ln[√0+1]+C)
=1-2+2ln[2]-2ln[1]
=2ln[2]-1

Answer 3

∫{1,4}[1/x(1+√x)]dx=∫{1,4}[1/x+x^(-1/2))]dx=[lnx+2x^(1/2)]{1,4}=(ln4+2√4)-(ln1+2√1)
=ln4+4-0-2=ln4+2

Answer 4

12121

Answer 5

2ln2-1