Question

积分 根号下(x^2+1)怎么算呀？谢谢！

Answer 1

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)|]+C
C为任意常数

Answer 2

设x=tant