$$$62 x + \left(12 x - 12\right) e^{2} - 62$$$ 的积分
您的输入
求$$$\int \left(62 x + \left(12 x - 12\right) e^{2} - 62\right)\, dx$$$。
解答
逐项积分：
$${\color{red}{\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x}}} = {\color{red}{\left(- \int{62 d x} + \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=62$$$：
$$\int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\int{62 d x}}} = \int{62 x d x} + \int{\left(12 x - 12\right) e^{2} d x} - {\color{red}{\left(62 x\right)}}$$
对 $$$c=62$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\int{62 x d x}}} = - 62 x + \int{\left(12 x - 12\right) e^{2} d x} + {\color{red}{\left(62 \int{x d x}\right)}}$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\int{x d x}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 62 x + \int{\left(12 x - 12\right) e^{2} d x} + 62 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
化简被积函数:
$$31 x^{2} - 62 x + {\color{red}{\int{\left(12 x - 12\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}}$$
对 $$$c=12 e^{2}$$$ 和 $$$f{\left(x \right)} = x - 1$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：
$$31 x^{2} - 62 x + {\color{red}{\int{12 \left(x - 1\right) e^{2} d x}}} = 31 x^{2} - 62 x + {\color{red}{\left(12 e^{2} \int{\left(x - 1\right)d x}\right)}}$$
逐项积分：
$$31 x^{2} - 62 x + 12 e^{2} {\color{red}{\int{\left(x - 1\right)d x}}} = 31 x^{2} - 62 x + 12 e^{2} {\color{red}{\left(- \int{1 d x} + \int{x d x}\right)}}$$
应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：
$$31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{\int{1 d x}}}\right) = 31 x^{2} - 62 x + 12 e^{2} \left(\int{x d x} - {\color{red}{x}}\right)$$
应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\int{x d x}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=31 x^{2} - 62 x + 12 e^{2} \left(- x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$
因此，
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = 31 x^{2} - 62 x + 12 \left(\frac{x^{2}}{2} - x\right) e^{2}$$
化简：
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)$$
加上积分常数：
$$\int{\left(62 x + \left(12 x - 12\right) e^{2} - 62\right)d x} = x \left(31 + 6 e^{2}\right) \left(x - 2\right)+C$$
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