$$$\left(4 x - 2\right) e^{x^{2} - x}$$$ 的积分

求$$$\int \left(4 x - 2\right) e^{x^{2} - x}\, dx$$$。

输入已重写为：$$$\int{\left(4 x - 2\right) e^{x^{2} - x} d x}=\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x}$$$。

设$$$u=x \left(x - 1\right)$$$。

则$$$du=\left(x \left(x - 1\right)\right)^{\prime }dx = \left(2 x - 1\right) dx$$$ (步骤见»)，并有$$$\left(2 x - 1\right) dx = du$$$。

积分变为

$${\color{red}{\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x}}} = {\color{red}{\int{2 e^{u} d u}}}$$

对 $$$c=2$$$ 和 $$$f{\left(u \right)} = e^{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{2 e^{u} d u}}} = {\color{red}{\left(2 \int{e^{u} d u}\right)}}$$

指数函数的积分为 $$$\int{e^{u} d u} = e^{u}$$$：

$$2 {\color{red}{\int{e^{u} d u}}} = 2 {\color{red}{e^{u}}}$$

回忆一下 $$$u=x \left(x - 1\right)$$$:

$$2 e^{{\color{red}{u}}} = 2 e^{{\color{red}{x \left(x - 1\right)}}}$$

因此，

$$\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x} = 2 e^{x \left(x - 1\right)}$$

加上积分常数：

$$\int{\left(4 x - 2\right) e^{x \left(x - 1\right)} d x} = 2 e^{x \left(x - 1\right)}+C$$

$$$\int \left(4 x - 2\right) e^{x^{2} - x}\, dx = 2 e^{x \left(x - 1\right)} + C$$$A