Question

求定积分 1/1+√xdx下限0上限1

Answer 1

令饥闹早羡√x=t
x=t^2
x=0,t=0,x=1,t=1
dx=2tdt
∫[0,1]1/(1+√x)dx
=∫[0,1]2tdt/(1+t)
=2∫烂睁罩[0,1] [1-1/(1+t)]dt
=2[t-ln(1+t)][0,1]
=2-2ln2