How would you find the indefinite integral of 8e^(3x+e^3x) dx

My first try was to use a subsitution and let u = e^3x then du= 3e^3x dx, but then i wasn't sure which e^3x would be u and was confused.

8 e^(3x + e^(3x))
= 8 e^(3x) e^(e^(3x))

Your substitution is correct. Always try to 'place' du first, then the rest becomes u:

u = e^(3x)
du = 3 e^(3x) dx

∫ 8 e^(3x + e^(3x)) dx = ∫ 8 e^(3x) e^(e^(3x)) dx
. . . . . . . . . . . . . . . . . . = 1/3 ∫ 8 e^(e^(3x)) * 3 e^(3x) dx
. . . . . . . . . . . . . . . . . . = 8/3 ∫ e^u du
. . . . . . . . . . . . . . . . . . = 8/3 e^u + C
. . . . . . . . . . . . . . . . . . = 8/3 e^(e^(3x)) + C

integral of 8e^(3x+e^3x) dx
= integral of 8e^(3x) * e^(e^3x) dx
= integral of (8/3) * e^(e^3x) de^(3x)
= (8/3) * e^(e^3x) + c