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?
 A: 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
$$\frac{1}{2\ln 7}7^{2x+3}+C.$$
A: You can always differentiate to be sure:
$$
\frac{\mathrm{d}}{\mathrm{d}x} \frac{7^{2x+3}}{2\ln 7} = \frac1{2 \ln 7} \frac{\mathrm{d}}{\mathrm{d}x} 7^{2x+3} = \frac1{2 \ln 7} (\ln 7) 7^{2x+3} \frac{\mathrm{d}}{\mathrm{d}x} (2x+3) = 7^{2x+3}
$$
Your result is sort of right; you're missing the constant and the differential after the substitution should be $\mathrm{d}u$, so the result is the following:
$$
\int 7^{2x+3}\ \mathrm{d}x = \frac1{2}\int 7^u \mathrm{d}u =\frac{7^{2x+3}}{2\ln 7} + C
$$
A: Yes it is correct.
except for the + Constant
A: You have, although after substitution it should be du not dx but I assume it is a typo and you should have a constant as well.
