令√(3x+9)=u,则:3x+9=u^2,∴x=(1/3)u^2-3,∴dx=(2/3)udu.
∴∫{e^[√(3x+9)]}dx
=(2/3)∫u·e^udu
=(2/3)∫ud(e^u)
=(2/3)u·e^u-(2/3)∫e^udu
=(2/3)√(3x+9)·e^[√(3x+9)]-(2/3)e^u+C
=(2/3)√(3x+9)·e^[√(3x+9)]-(2/3)e^[√(3x+9)]+C.
令√(3x+9)=u,则:3x+9=u^2,∴x=(1/3)u^2-3,∴dx=(2/3)udu.
∴∫{e^[√(3x+9)]}dx
=(2/3)∫u·e^udu
=(2/3)∫ud(e^u)
=(2/3)u·e^u-(2/3)∫e^udu
=(2/3)√(3x+9)·e^[√(3x+9)]-(2/3)e^u+C
=(2/3)√(3x+9)·e^[√(3x+9)]-(2/3)e^[√(3x+9)]+C.