$$$\ln\left(x e^{8} - 9\right)$$$ 的积分

求$$$\int \ln\left(x e^{8} - 9\right)\, dx$$$。

设$$$u=x e^{8} - 9$$$。

则$$$du=\left(x e^{8} - 9\right)^{\prime }dx = e^{8} dx$$$ (步骤见»)，并有$$$dx = \frac{du}{e^{8}}$$$。

该积分可以改写为

$${\color{red}{\int{\ln{\left(x e^{8} - 9 \right)} d x}}} = {\color{red}{\int{\frac{\ln{\left(u \right)}}{e^{8}} d u}}}$$

对 $$$c=e^{-8}$$$ 和 $$$f{\left(u \right)} = \ln{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{\ln{\left(u \right)}}{e^{8}} d u}}} = {\color{red}{\frac{\int{\ln{\left(u \right)} d u}}{e^{8}}}}$$

对于积分$$$\int{\ln{\left(u \right)} d u}$$$，使用分部积分法$$$\int \operatorname{\theta} \operatorname{dv} = \operatorname{\theta}\operatorname{v} - \int \operatorname{v} \operatorname{d\theta}$$$。

设 $$$\operatorname{\theta}=\ln{\left(u \right)}$$$ 和 $$$\operatorname{dv}=du$$$。

则 $$$\operatorname{d\theta}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d u}=u$$$ (步骤见 »)。

所以，

$$\frac{{\color{red}{\int{\ln{\left(u \right)} d u}}}}{e^{8}}=\frac{{\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}}{e^{8}}=\frac{{\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}}{e^{8}}$$

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$\frac{u \ln{\left(u \right)} - {\color{red}{\int{1 d u}}}}{e^{8}} = \frac{u \ln{\left(u \right)} - {\color{red}{u}}}{e^{8}}$$

回忆一下 $$$u=x e^{8} - 9$$$:

$$\frac{- {\color{red}{u}} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}}{e^{8}} = \frac{- {\color{red}{\left(x e^{8} - 9\right)}} + {\color{red}{\left(x e^{8} - 9\right)}} \ln{\left({\color{red}{\left(x e^{8} - 9\right)}} \right)}}{e^{8}}$$

因此，

$$\int{\ln{\left(x e^{8} - 9 \right)} d x} = \frac{- x e^{8} + \left(x e^{8} - 9\right) \ln{\left(x e^{8} - 9 \right)} + 9}{e^{8}}$$

化简：

$$\int{\ln{\left(x e^{8} - 9 \right)} d x} = \frac{\left(x e^{8} - 9\right) \left(\ln{\left(x e^{8} - 9 \right)} - 1\right)}{e^{8}}$$

加上积分常数：

$$\int{\ln{\left(x e^{8} - 9 \right)} d x} = \frac{\left(x e^{8} - 9\right) \left(\ln{\left(x e^{8} - 9 \right)} - 1\right)}{e^{8}}+C$$

$$$\int \ln\left(x e^{8} - 9\right)\, dx = \frac{\left(x e^{8} - 9\right) \left(\ln\left(x e^{8} - 9\right) - 1\right)}{e^{8}} + C$$$A