Integral of $$$e^{x} \tan^{x}{\left(e \right)}$$$
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Your Input
Find $$$\int e^{x} \tan^{x}{\left(e \right)}\, dx$$$.
Solution
The input is rewritten: $$$\int{e^{x} \tan^{x}{\left(e \right)} d x}=\int{\left(e \tan{\left(e \right)}\right)^{x} d x}$$$.
Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=e \tan{\left(e \right)}$$$:
$${\color{red}{\int{\left(e \tan{\left(e \right)}\right)^{x} d x}}} = {\color{red}{\frac{\left(e \tan{\left(e \right)}\right)^{x}}{\ln{\left(e \tan{\left(e \right)} \right)}}}}$$
Therefore,
$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{\left(e \tan{\left(e \right)}\right)^{x}}{\ln{\left(e \tan{\left(e \right)} \right)}}$$
Simplify:
$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln{\left(- \tan{\left(e \right)} \right)} + 1 + i \pi}$$
Add the constant of integration:
$$\int{\left(e \tan{\left(e \right)}\right)^{x} d x} = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln{\left(- \tan{\left(e \right)} \right)} + 1 + i \pi}+C$$
Answer
$$$\int e^{x} \tan^{x}{\left(e \right)}\, dx = \frac{e^{x} \tan^{x}{\left(e \right)}}{\ln\left(- \tan{\left(e \right)}\right) + 1 + i \pi} + C$$$A