$$$\frac{2}{7 - x^{2}}$$$ 的积分

求$$$\int \frac{2}{7 - x^{2}}\, dx$$$。

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \frac{1}{7 - x^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{2}{7 - x^{2}} d x}}} = {\color{red}{\left(2 \int{\frac{1}{7 - x^{2}} d x}\right)}}$$

进行部分分式分解（步骤可见»）:

$$2 {\color{red}{\int{\frac{1}{7 - x^{2}} d x}}} = 2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}}$$

逐项积分：

$$2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}} = 2 {\color{red}{\left(- \int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x} + \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}\right)}}$$

对 $$$c=\frac{\sqrt{7}}{14}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - \sqrt{7}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x}}} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x - \sqrt{7}} d x}}{14}\right)}}$$

设$$$u=x - \sqrt{7}$$$。

则$$$du=\left(x - \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

因此，

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x - \sqrt{7}} d x}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

回忆一下 $$$u=x - \sqrt{7}$$$:

$$- \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} = - \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x - \sqrt{7}\right)}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}$$

对 $$$c=\frac{\sqrt{7}}{14}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x + \sqrt{7}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}}} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x + \sqrt{7}} d x}}{14}\right)}}$$

设$$$u=x + \sqrt{7}$$$。

则$$$du=\left(x + \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

该积分可以改写为

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x + \sqrt{7}} d x}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

回忆一下 $$$u=x + \sqrt{7}$$$:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x + \sqrt{7}\right)}}}\right| \right)}}{7}$$

因此，

$$\int{\frac{2}{7 - x^{2}} d x} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{x + \sqrt{7}}\right| \right)}}{7}$$

化简：

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}$$

加上积分常数：

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}+C$$

$$$\int \frac{2}{7 - x^{2}}\, dx = \frac{\sqrt{7} \left(- \ln\left(\left|{x - \sqrt{7}}\right|\right) + \ln\left(\left|{x + \sqrt{7}}\right|\right)\right)}{7} + C$$$A