根号(x的平方+4)的积分怎么求
∫[√(x²+4)]dx=2∫[√(x/2)²+1]dx
令x/2=tanu,则x=2tanu,dx=2du/cos²u=2sec²udu
故原式=4∫sec³udu=4∫secud(tanu)=4[secutanu-∫tanud(secu)]=4[secutanu-∫tan²usecudu]
=4[secutanu-∫(sec²u-1)secudu]=4secutanu-4∫sec³udu+4∫secudu
故有8∫sec³udu=4secutanu+4ln(secu+tanu)+C
∴4∫sec³udu=2secutanu+2ln(secu+tanu)+C
用tanu=x/2,secu=[√(x²+4)]/2代入,
于是得：∫[√(x²+4)]dx=2∫[√(x/2)²+1]dx=4∫sec³udu=2secutanu+2ln(secu+tanu)+C
=(x/2)√(x²+4)+2ln{[x+√(x²+4)]/2}+C
