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.

Our integral is:
$$\int 6^{-2x} \ dx$$
Rewrite the integral:
$$\int \left(6^{-2}\right)^x \ dx$$
$$=\int \left(\frac{1}{36}\right)^x \ dx$$
Remember that:
$$\int a^x \ dx=\frac{a^x}{\ln a}+C$$
In this case our $a$ is $\frac{1}{36}$
$$\int \left(\frac 1{36}\right)^x \ dx=\frac{\left(\frac{1}{36}\right)^x}{\ln \left(\frac{1}{36}\right)}+C$$
$$=\frac{6^{-2x}}{\ln\left(6^{-2}\right)}+C$$
$$=-\frac{6^{-2x}}{2 \ln 6}+C$$
Hope I helped!
