Evaluating $\int 7^{2x+3} dx$ I have this integral to evaluate: $\int 7^{2x+3} dx$

u substitution should work and you are left with $\frac{1}{2}\int7^udx$

And the final answer should be: $$\frac{7^{2x+3}}{2\ln7}$$

I wasn't too sure about this... so did I do this correctly?
You do not show detail, but you presumably found $\int 7^u\,du$ by looking up a formula for $\int a^u\,du$. That is perfectly correct. But I  would use a slightly different approach, which is a bit more complicated but does not rely on remembering $\int a^u\,du$.

Note that 
$$7^{2x+3}=(e^{\ln 7})^{2x+3}=e^{(\ln7)(2x+3)}. \qquad\qquad (\ast)$$
Let $v=(\ln 7)(2x+3)$. Then $dv=(\ln 7)(2) \,dx$. Substituting, we find that
$$\int e^{(\ln7)(2x+3)}\,dx=\int \frac{1}{2\ln 7}e^v\,dv=\frac{1}{2\ln 7}e^v+C.$$
Finally, by $(\ast)$, $e^v=7^{2x+3}$ so our integral is
$$\frac{1}{2\ln 7}7^{2x+3}+C.$$
Yes it is correct.

except for the + Constant