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$$$\sin{\left(x \right)} - \frac{1}{x}$$$の積分
入力内容
$$$\int \left(\sin{\left(x \right)} - \frac{1}{x}\right)\, dx$$$ を求めよ。
解答
項別に積分せよ:
$${\color{red}{\int{\left(\sin{\left(x \right)} - \frac{1}{x}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x} d x} + \int{\sin{\left(x \right)} d x}\right)}}$$
$$$\frac{1}{x}$$$ の不定積分は $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$ です:
$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{x} d x}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$
正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:
$$- \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{\sin{\left(x \right)} d x}}} = - \ln{\left(\left|{x}\right| \right)} + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$
したがって、
$$\int{\left(\sin{\left(x \right)} - \frac{1}{x}\right)d x} = - \ln{\left(\left|{x}\right| \right)} - \cos{\left(x \right)}$$
積分定数を加える:
$$\int{\left(\sin{\left(x \right)} - \frac{1}{x}\right)d x} = - \ln{\left(\left|{x}\right| \right)} - \cos{\left(x \right)}+C$$
解答
$$$\int \left(\sin{\left(x \right)} - \frac{1}{x}\right)\, dx = \left(- \ln\left(\left|{x}\right|\right) - \cos{\left(x \right)}\right) + C$$$A