$$$\frac{1}{9 x e^{2} - 4}$$$ 的积分

求$$$\int \frac{1}{9 x e^{2} - 4}\, dx$$$。

设$$$u=9 x e^{2} - 4$$$。

则$$$du=\left(9 x e^{2} - 4\right)^{\prime }dx = 9 e^{2} dx$$$ (步骤见»)，并有$$$dx = \frac{du}{9 e^{2}}$$$。

积分变为

$${\color{red}{\int{\frac{1}{9 x e^{2} - 4} d x}}} = {\color{red}{\int{\frac{1}{9 u e^{2}} d u}}}$$

对 $$$c=\frac{1}{9 e^{2}}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{1}{9 u e^{2}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{9 e^{2}}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{9 e^{2}} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{9 e^{2}}$$

回忆一下 $$$u=9 x e^{2} - 4$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{9 e^{2}} = \frac{\ln{\left(\left|{{\color{red}{\left(9 x e^{2} - 4\right)}}}\right| \right)}}{9 e^{2}}$$

因此，

$$\int{\frac{1}{9 x e^{2} - 4} d x} = \frac{\ln{\left(\left|{9 x e^{2} - 4}\right| \right)}}{9 e^{2}}$$

加上积分常数：

$$\int{\frac{1}{9 x e^{2} - 4} d x} = \frac{\ln{\left(\left|{9 x e^{2} - 4}\right| \right)}}{9 e^{2}}+C$$

$$$\int \frac{1}{9 x e^{2} - 4}\, dx = \frac{\ln\left(\left|{9 x e^{2} - 4}\right|\right)}{9 e^{2}} + C$$$A