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