求定积分∫dx/[根号(1+x^2)^3],上限1,下限0.

令x＝tant,dx＝(sect)^2dt。x＝0时t＝0,x＝1时,t＝π/4,所以∫(0,1) dx/√[(1+x^2)^3]＝∫(0,π/4) cost dt＝sin(π/4)＝√2/2