$$$\frac{x + 4}{x^{2} + 4 x + 4}$$$ 的积分

求$$$\int \frac{x + 4}{x^{2} + 4 x + 4}\, dx$$$。

将线性项改写为 $$$x + 4=x\color{red}{+2-2}+4=x+2+2$$$，并将表达式拆分:

$${\color{red}{\int{\frac{x + 4}{x^{2} + 4 x + 4} d x}}} = {\color{red}{\int{\left(\frac{x + 2}{x^{2} + 4 x + 4} + \frac{2}{x^{2} + 4 x + 4}\right)d x}}}$$

逐项积分：

$${\color{red}{\int{\left(\frac{x + 2}{x^{2} + 4 x + 4} + \frac{2}{x^{2} + 4 x + 4}\right)d x}}} = {\color{red}{\left(\int{\frac{x + 2}{x^{2} + 4 x + 4} d x} + \int{\frac{2}{x^{2} + 4 x + 4} d x}\right)}}$$

设$$$u=x^{2} + 4 x + 4$$$。

则$$$du=\left(x^{2} + 4 x + 4\right)^{\prime }dx = \left(2 x + 4\right) dx$$$ (步骤见»)，并有$$$\left(2 x + 4\right) dx = du$$$。

因此，

$$\int{\frac{2}{x^{2} + 4 x + 4} d x} + {\color{red}{\int{\frac{x + 2}{x^{2} + 4 x + 4} d x}}} = \int{\frac{2}{x^{2} + 4 x + 4} d x} + {\color{red}{\int{\frac{1}{2 u} d u}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

$$\int{\frac{2}{x^{2} + 4 x + 4} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \int{\frac{2}{x^{2} + 4 x + 4} d x} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=x^{2} + 4 x + 4$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + \int{\frac{2}{x^{2} + 4 x + 4} d x} = \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} + 4 x + 4\right)}}}\right| \right)}}{2} + \int{\frac{2}{x^{2} + 4 x + 4} d x}$$

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2} + 4 x + 4}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + {\color{red}{\int{\frac{2}{x^{2} + 4 x + 4} d x}}} = \frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + {\color{red}{\left(2 \int{\frac{1}{x^{2} + 4 x + 4} d x}\right)}}$$

配平方（步骤见»）：$$$x^{2} + 4 x + 4 = \left(x + 2\right)^{2}$$$:

$$\frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{1}{x^{2} + 4 x + 4} d x}}} = \frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{1}{\left(x + 2\right)^{2}} d x}}}$$

设$$$u=x + 2$$$。

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

该积分可以改写为

$$\frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{1}{\left(x + 2\right)^{2}} d x}}} = \frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

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

回忆一下 $$$u=x + 2$$$:

$$\frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} - 2 {\color{red}{u}}^{-1} = \frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} - 2 {\color{red}{\left(x + 2\right)}}^{-1}$$

因此，

$$\int{\frac{x + 4}{x^{2} + 4 x + 4} d x} = \frac{\ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)}}{2} - \frac{2}{x + 2}$$

化简：

$$\int{\frac{x + 4}{x^{2} + 4 x + 4} d x} = \frac{\left(x + 2\right) \ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)} - 4}{2 \left(x + 2\right)}$$

加上积分常数：

$$\int{\frac{x + 4}{x^{2} + 4 x + 4} d x} = \frac{\left(x + 2\right) \ln{\left(\left|{x^{2} + 4 x + 4}\right| \right)} - 4}{2 \left(x + 2\right)}+C$$

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