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
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