[tex]\int {3^{x+3^x}} \, dx = \int {3^x * 3^{3^x}} \, dx[/tex]
From here, make a u-substitution, where [tex]u = 3^x[/tex] and [tex]du = (\ln 3)3^x \, dx[/tex]. Then [tex]dx = \frac{1}{(\ln 3)3^x} du[/tex], and:
[tex]\int {3^x * 3^{3^x}} \, dx[/tex]
[tex]= \int {\frac{3^u}{\ln 3}} \, du[/tex]
[tex]= \frac{1}{\ln 3} \int {3^u} \, du[/tex]
[tex]= \frac{1}{\ln 3}(\frac{3^u}{\ln 3}) + C[/tex]
[tex]= \frac{3^u}{\ln^2 3} + C[/tex]
[tex]= \bf \frac{3^{3^x}}{\ln^2 3} + C[/tex]
