积分基础：$\int \ln x\,dx$ 与 $\in

1. 分部积分：取 $u=\ln x,\ dv=dx$，得

$$\int \ln x\,dx=x\ln x-x+C.$$

2. 利用公式 $\int \sec x\,dx=\ln|\sec x+\tan x|+C$，因此

$$\int_0^{\pi/6}\sec x\,dx=\ln\bigl(\sec(\tfrac{\pi}{6})+\tan(\tfrac{\pi}{6})\bigr)-\ln(1)=\tfrac12\ln 3.$$

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Original Explanation

By parts. Let $u=\ln x,\; dv=dx,$ so $du=\tfrac{dx}{x},\;v=x.$
$$\int \ln x\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.$$

• $\frac{d}{dx}\sec x=\sec x\tan x,\;\frac{d}{dx}\tan x=\sec^2 x.$
• $\frac{d}{dx}\ln|\sec x+\tan x|=\sec x.$

Hence $$\int \sec x\,dx = \ln|\sec x+\tan x|+C.$$ Thus $$\int_0^{\pi/6}\sec x\,dx = \Bigl[\ln|\sec x+\tan x|\Bigr]_0^{\pi/6} = \ln\bigl|\sec(\tfrac{\pi}{6})+\tan(\tfrac{\pi}{6})\bigr| - \ln\bigl|\sec 0+\tan 0\bigr| = \ln(\sqrt{3}) = \tfrac12\ln 3.$$