$$$2^{- \frac{t}{5}}$$$の積分

$$$\int 2^{- \frac{t}{5}}\, dt$$$ を求めよ。

入力は次のように書き換えられます: $$$\int{2^{- \frac{t}{5}} d t}=\int{\left(\frac{2^{\frac{4}{5}}}{2}\right)^{t} d t}$$$。

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

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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