Integral of $$$2^{- \frac{t}{5}}$$$

Find $$$\int 2^{- \frac{t}{5}}\, dt$$$.

The input is rewritten: $$$\int{2^{- \frac{t}{5}} d t}=\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t}$$$.

Apply the exponential rule $$$\int{a^{t} d t} = \frac{a^{t}}{\ln{\left(a \right)}}$$$ with $$$a=\frac{2^{\frac{4}{5}}}{2}$$$:

$${\color{red}{\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t}}} = {\color{red}{\frac{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t}}{\ln{\left(\frac{2^{\frac{4}{5}}}{2} \right)}}}}$$

Therefore,

$$\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t} = \frac{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t}}{\ln{\left(\frac{2^{\frac{4}{5}}}{2} \right)}}$$

Simplify:

$$\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t} = - \frac{5 \cdot 2^{- \frac{t}{5}}}{\ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t} = - \frac{5 \cdot 2^{- \frac{t}{5}}}{\ln{\left(2 \right)}}+C$$

$$$\int 2^{- \frac{t}{5}}\, dt = - \frac{5 \cdot 2^{- \frac{t}{5}}}{\ln\left(2\right)} + C$$$A