Integrating $3^{2x}$ ($\int 3^{2x} dx $) I am tring to integrate the following: $3^{2x}$
My working has been shown as the following
$$\int 3^{2x} dx $$
I used the substitution $u = 2x$, so 
$$\frac{du}{2} = dx $$
hence
$$\int 3^{2x} dx = \int 3^{u} \frac{du}{2} = \frac{1}{2}\int 3^{u} du $$ 
However, I can't go any further.
From this, I can't seem to prove the general result of  $\int a^{x} dx $ either.
Can someone help?
Thanks!

Formula -
$$\int a^x = \frac{a^x}{\ln a} + c$$ 
Now we have,
$$\frac{1}{2}\int 3^{u} du $$ 
$$=  \frac{1}{2} \cdot \frac{3^u}{\ln 3} + c$$ 
$$=  \frac{1}{2} \cdot \frac{3^{2x}}{\ln 3} + c$$ 
