令√[(1-x)/(1+x)]=t
(1-x)/(1+x)=t^2
1-x=t^2+t^2x
(1+t^2)x=1-t^2
x=(1-t^2)/(1+t^2)
dx=[(1-t^2)'(1+t^2)-(1-t^2)(1+t^2)']/(1+t^2)^2 dt
=[-2t(1+t^2)-2t(1-t^2)]/(1+t^2)^2 dt
=-4tdt/(1+t^2)^2
=-2d(1+t^2)/(1+t^2)^2
=2d[1/(1+t^2)]
当x=0时 t=1
当x=1时 t=0
∴原化化为
∫(1->0)t2d[1/(1+t^2)]
=2t/(1+t^2)|(1->0)-2∫(1->0)1/(1+t^2)*dt
=2t/(1+t^2)|(1->0)-2arctant|(1->0)
=2(0-1/2)-2(0-π/4)
=π/2-1