$$$2^{5 x} 5^{- 2 x}$$$の積分

$$$\int 2^{5 x} 5^{- 2 x}\, dx$$$ を求めよ。

入力は次のように書き換えられます: $$$\int{2^{5 x} 5^{- 2 x} d x}=\int{\left(\frac{32}{25}\right)^{x} d x}$$$。

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=\frac{32}{25}$$$:

$${\color{red}{\int{\left(\frac{32}{25}\right)^{x} d x}}} = {\color{red}{\frac{\left(\frac{32}{25}\right)^{x}}{\ln{\left(\frac{32}{25} \right)}}}}$$

したがって、

$$\int{\left(\frac{32}{25}\right)^{x} d x} = \frac{\left(\frac{32}{25}\right)^{x}}{\ln{\left(\frac{32}{25} \right)}}$$

簡単化せよ:

$$\int{\left(\frac{32}{25}\right)^{x} d x} = \frac{\left(\frac{32}{25}\right)^{x}}{- 2 \ln{\left(5 \right)} + 5 \ln{\left(2 \right)}}$$

積分定数を加える:

$$\int{\left(\frac{32}{25}\right)^{x} d x} = \frac{\left(\frac{32}{25}\right)^{x}}{- 2 \ln{\left(5 \right)} + 5 \ln{\left(2 \right)}}+C$$

$$$\int 2^{5 x} 5^{- 2 x}\, dx = \frac{\left(\frac{32}{25}\right)^{x}}{- 2 \ln\left(5\right) + 5 \ln\left(2\right)} + C$$$A