$$$\frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}$$$ 的积分

求$$$\int \frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}\, dx$$$。

设$$$u=\ln{\left(x \right)}$$$。

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

该积分可以改写为

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

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha= u $$$:

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

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

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

逐项积分：

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

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

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

设$$$v=2 u$$$。

则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»)，并有$$$du = \frac{dv}{2}$$$。

因此，

$$\frac{u}{2} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{2}$$

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

$$\frac{u}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{2}$$

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

$$\frac{u}{2} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{4} = \frac{u}{2} + \frac{{\color{red}{\sin{\left(v \right)}}}}{4}$$

回忆一下 $$$v=2 u$$$:

$$\frac{u}{2} + \frac{\sin{\left({\color{red}{v}} \right)}}{4} = \frac{u}{2} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$\frac{\sin{\left(2 {\color{red}{u}} \right)}}{4} + \frac{{\color{red}{u}}}{2} = \frac{\sin{\left(2 {\color{red}{\ln{\left(x \right)}}} \right)}}{4} + \frac{{\color{red}{\ln{\left(x \right)}}}}{2}$$

因此，

$$\int{\frac{1}{x \sec^{2}{\left(\ln{\left(x \right)} \right)}} d x} = \frac{\ln{\left(x \right)}}{2} + \frac{\sin{\left(2 \ln{\left(x \right)} \right)}}{4}$$

加上积分常数：

$$\int{\frac{1}{x \sec^{2}{\left(\ln{\left(x \right)} \right)}} d x} = \frac{\ln{\left(x \right)}}{2} + \frac{\sin{\left(2 \ln{\left(x \right)} \right)}}{4}+C$$

$$$\int \frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}\, dx = \left(\frac{\ln\left(x\right)}{2} + \frac{\sin{\left(2 \ln\left(x\right) \right)}}{4}\right) + C$$$A