Evaluate the integral: integral of $\int 6^{-2x} dx$ $$\displaystyle \int6^{-2x}dx$$
I got $-\dfrac{6^{-2x}}{2\ln6}$  but I'm not confident that its correct.

Let $y=6^{-2x}$, then
\begin{align}
\ln y=\ln\left(6^{-2x}\right)=-2x\ln6\qquad\rightarrow\qquad y=e^{-(2\ln6)x}
\end{align}
Hence
\begin{align}
\int6^{-2x}\,dx&=\int e^{-(2\ln6)x}\,dx
\end{align}
Remember
\begin{align}
\int e^{ax}\,dx=\frac{e^{ax}}{a}+C
\end{align}
then
\begin{align}
\int6^{-2x}\,dx&=\int e^{-(2\ln6)x}\,dx\\
&=\frac{e^{-(2\ln6)x}}{-(2\ln6)}+C\\
&=-\frac{e^{-(2\ln6)x}}{2\ln6}+C\\
&=-\frac{6^{-2x}}{2\ln6}+C\\
\end{align}
