$$$\frac{e^{x} - 1}{x}$$$の積分
入力内容
$$$\int \frac{e^{x} - 1}{x}\, dx$$$ を求めよ。
解答
Expand the expression:
$${\color{red}{\int{\frac{e^{x} - 1}{x} d x}}} = {\color{red}{\int{\left(\frac{e^{x}}{x} - \frac{1}{x}\right)d x}}}$$
項別に積分せよ:
$${\color{red}{\int{\left(\frac{e^{x}}{x} - \frac{1}{x}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x} d x} + \int{\frac{e^{x}}{x} d x}\right)}}$$
$$$\frac{1}{x}$$$ の不定積分は $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$ です:
$$\int{\frac{e^{x}}{x} d x} - {\color{red}{\int{\frac{1}{x} d x}}} = \int{\frac{e^{x}}{x} d x} - {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$
この積分（指数積分）には閉形式はありません:
$$- \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{\frac{e^{x}}{x} d x}}} = - \ln{\left(\left|{x}\right| \right)} + {\color{red}{\operatorname{Ei}{\left(x \right)}}}$$
したがって、
$$\int{\frac{e^{x} - 1}{x} d x} = - \ln{\left(\left|{x}\right| \right)} + \operatorname{Ei}{\left(x \right)}$$
積分定数を加える:
$$\int{\frac{e^{x} - 1}{x} d x} = - \ln{\left(\left|{x}\right| \right)} + \operatorname{Ei}{\left(x \right)}+C$$
解答
$$$\int \frac{e^{x} - 1}{x}\, dx = \left(- \ln\left(\left|{x}\right|\right) + \operatorname{Ei}{\left(x \right)}\right) + C$$$A