Does look right? As far as correct form, etc..?

$\displaystyle \int_{1}^{5} 3^{2x} dx$.

Using $\displaystyle \int a^{u} du = \dfrac{a^{u}}{\ln a} + C$ and $\displaystyle \dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a$.

$\displaystyle u = 2x$.

$\displaystyle du = 2$.

$\displaystyle a = 3$.

$\displaystyle \dfrac{1}{2} \int_{1}^{5} 3^{u} du$.

$\displaystyle \int_{1}^{5} \dfrac{1}{2} 3^{u} du $.

$\displaystyle \int_{1}^{5} \dfrac{3^{u}}{2} du $.

$\displaystyle \dfrac{3^{2(u)}}{2\ln 3} |_{1}^{5} $.

$\displaystyle \dfrac{3^{2(5)}}{2\ln 3}$ - $\displaystyle \dfrac{3^{2(1)}}{2\ln 3}$.

$\displaystyle \dfrac{3^{10}}{2\ln 3}$ - $\displaystyle \dfrac{3^{2}}{2\ln 3}$.

$\displaystyle \dfrac{59049}{2\ln 3}$ - $\displaystyle \dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}$