$$$\frac{r \sin{\left(\ln\left(x\right) \right)}}{x}$$$ 关于$$$x$$$的积分

求$$$\int \frac{r \sin{\left(\ln\left(x\right) \right)}}{x}\, dx$$$。

对 $$$c=r$$$ 和 $$$f{\left(x \right)} = \frac{\sin{\left(\ln{\left(x \right)} \right)}}{x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}} = {\color{red}{r \int{\frac{\sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}}$$

设$$$u=\ln{\left(x \right)}$$$。

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

因此，

$$r {\color{red}{\int{\frac{\sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}} = r {\color{red}{\int{\sin{\left(u \right)} d u}}}$$

正弦函数的积分为 $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$r {\color{red}{\int{\sin{\left(u \right)} d u}}} = r {\color{red}{\left(- \cos{\left(u \right)}\right)}}$$

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$- r \cos{\left({\color{red}{u}} \right)} = - r \cos{\left({\color{red}{\ln{\left(x \right)}}} \right)}$$

因此，

$$\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x} = - r \cos{\left(\ln{\left(x \right)} \right)}$$

加上积分常数：

$$\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x} = - r \cos{\left(\ln{\left(x \right)} \right)}+C$$

$$$\int \frac{r \sin{\left(\ln\left(x\right) \right)}}{x}\, dx = - r \cos{\left(\ln\left(x\right) \right)} + C$$$A