积分 根号下(x^2+1)怎么算呀?

这个东西挺麻烦的,耐心看完设I=∫√(x²+1) dx则I=x√(x²+1)-∫xd[√(x²+1)]=x√(x²+1)-∫[x²/√(x²+1)]dx=x√(x²+1)-∫[(x²+1)/√(x²+1)]dx+∫[1/√(x²+1)]dx=x√(x²+1)-I+∫[1/√(x²+1)]dx∴I=(1/2){x√(x²+1)+∫[1/√(x²+1)]dx}求∫[1/√(x²+1)]dx：设x=tant,则√(x²+1)=sect,dx=sec²tdt∫[1/√(x²+1)]dx=∫sec²t/sect dt=∫sect dt=ln|tant+sect|+C=ln|x+√(x²+1)|+C∴I=(1/2){x√(x²+1)+∫[1/√(x²+1)]dx}=(1/2)[x√(x²+1)+ln|x+√(x²+1)|]+CC为任意常数