$$$e^{x} \tan^{x}{\left(e \right)}$$$の積分

$$$\int e^{x} \tan^{x}{\left(e \right)}\, dx$$$ を求めよ。

入力は次のように書き換えられます: $$$\int{e^{x} \tan^{x}{\left(e \right)} d x}=\int{\left(e \tan{\left(e \right)}\right)^{x} d x}$$$。

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

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

したがって、

$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{\left(e \tan{\left(e \right)}\right)^{x}}{\ln{\left(e \tan{\left(e \right)} \right)}}$$

簡単化せよ:

$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln{\left(- \tan{\left(e \right)} \right)} + 1 + i \pi}$$

積分定数を加える:

$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln{\left(- \tan{\left(e \right)} \right)} + 1 + i \pi}+C$$

$$$\int e^{x} \tan^{x}{\left(e \right)}\, dx = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln\left(- \tan{\left(e \right)}\right) + 1 + i \pi} + C$$$A