這個(gè)東西挺麻煩的,耐心看完
設(shè)I=∫√(x²+1) dx
則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：
設(shè)x=tant,則√(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為任意常數(shù)