Question

How to integrate 3^(x+3^x)

Answer 1

$\int {3^{x+3^x}} \, dx = \int {3^x * 3^{3^x}} \, dx$

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:

$\int {3^x * 3^{3^x}} \, dx$
$= \int {\frac{3^u}{\ln 3}} \, du$
$= \frac{1}{\ln 3} \int {3^u} \, du$
$= \frac{1}{\ln 3}(\frac{3^u}{\ln 3}) + C$
$= \frac{3^u}{\ln^2 3} + C$
$= \bf \frac{3^{3^x}}{\ln^2 3} + C$