Integral of $$$11^{- k} 4^{k}$$$

Find $$$\int 11^{- k} 4^{k}\, dk$$$.

The input is rewritten: $$$\int{11^{- k} 4^{k} d k}=\int{\left(\frac{4}{11}\right)^{k} d k}$$$.

Apply the exponential rule $$$\int{a^{k} d k} = \frac{a^{k}}{\ln{\left(a \right)}}$$$ with $$$a=\frac{4}{11}$$$:

$${\color{red}{\int{\left(\frac{4}{11}\right)^{k} d k}}} = {\color{red}{\frac{\left(\frac{4}{11}\right)^{k}}{\ln{\left(\frac{4}{11} \right)}}}}$$

Therefore,

$$\int{\left(\frac{4}{11}\right)^{k} d k} = \frac{\left(\frac{4}{11}\right)^{k}}{\ln{\left(\frac{4}{11} \right)}}$$

Simplify:

$$\int{\left(\frac{4}{11}\right)^{k} d k} = \frac{\left(\frac{4}{11}\right)^{k}}{- \ln{\left(11 \right)} + 2 \ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{\left(\frac{4}{11}\right)^{k} d k} = \frac{\left(\frac{4}{11}\right)^{k}}{- \ln{\left(11 \right)} + 2 \ln{\left(2 \right)}}+C$$

$$$\int 11^{- k} 4^{k}\, dk = \frac{\left(\frac{4}{11}\right)^{k}}{- \ln\left(11\right) + 2 \ln\left(2\right)} + C$$$A