$$$62 x + \left(12 x - 12\right) e^{2} - 62$$$ 的積分
您的輸入
求$$$\int \left(62 x + \left(12 x - 12\right) e^{2} - 62\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x}}} = {\color{red}{\left(- \int{62 d x} + \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x}\right)}}$$
配合 $$$c=62$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$\int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\int{62 d x}}} = \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\left(62 x\right)}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=62$$$ 與 $$$f{\left(x \right)} = x$$$：
$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\int{62 x d x}}} = - 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\left(62 \int{x d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：
$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\int{x d x}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
簡化被積函數:
$$31 x^{2} - 62 x + {\color{red}{\int{\left(12 x - 12\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}}$$
套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=12 e^{2}$$$ 與 $$$f{\left(x \right)} = x - 1$$$：
$$31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\left(12 e^{2} \int{\left(x - 1\right)d x}\right)}}$$
逐項積分：
$$31 x^{2} - 62 x + 12 e^{2} {\color{red}{\int{\left(x - 1\right)d x}}} = 31 x^{2} - 62 x + 12 e^{2} {\color{red}{\left(- \int{1 d x} + \int{x d x}\right)}}$$
配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{\int{1 d x}}}\right) = 31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{x}}\right)$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：
$$31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\int{x d x}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$
因此，
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = 31 x^{2} - 62 x + 12 \left(\frac{x^{2}}{2} - x\right) e^{2}$$
化簡：
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)$$
加上積分常數：
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)+C$$
答案
$$$\int \left(62 x + \left(12 x - 12\right) e^{2} - 62\right)\, dx = x \left(31 + 6 e^{2}\right) \left(x - 2\right) + C$$$A