$$$- \frac{3}{\left(x + 2\right)^{2}}$$$ 的積分

求$$$\int \left(- \frac{3}{\left(x + 2\right)^{2}}\right)\, dx$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=-3$$$ 與 $$$f{\left(x \right)} = \frac{1}{\left(x + 2\right)^{2}}$$$：

$${\color{red}{\int{\left(- \frac{3}{\left(x + 2\right)^{2}}\right)d x}}} = {\color{red}{\left(- 3 \int{\frac{1}{\left(x + 2\right)^{2}} d x}\right)}}$$

令 $$$u=x + 2$$$。

則 $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

$$- 3 {\color{red}{\int{\frac{1}{\left(x + 2\right)^{2}} d x}}} = - 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

$$- 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- 3 {\color{red}{\int{u^{-2} d u}}}=- 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- 3 {\color{red}{\left(- u^{-1}\right)}}=- 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

回顧一下 $$$u=x + 2$$$：

$$3 {\color{red}{u}}^{-1} = 3 {\color{red}{\left(x + 2\right)}}^{-1}$$

因此，

$$\int{\left(- \frac{3}{\left(x + 2\right)^{2}}\right)d x} = \frac{3}{x + 2}$$

加上積分常數：

$$\int{\left(- \frac{3}{\left(x + 2\right)^{2}}\right)d x} = \frac{3}{x + 2}+C$$

$$$\int \left(- \frac{3}{\left(x + 2\right)^{2}}\right)\, dx = \frac{3}{x + 2} + C$$$A