치환적분법

Setting $u=\log x$, we have $\mathrm{d}u = \frac{1}{x}\,\mathrm{d}x$ and so

$$\int_{e}^{e^4} \frac{ \mathrm{d}x}{{blue}x\cdot \log x} = \int_{x=e}^{x=e^4} \frac1{ \log x} \cdot \frac{1}{x}\,\mathrm{d}x = \int_{u=1}^{u=4} \frac{1}{{red}u}\, \mathrm{d}u,$$

since $u=\log(e)=1$ when $x=e$ and $u=\log(e^4)=4$ when $x=e^4$. Then, by the Fundamental Theorem of Calculus Part 2,

$$\int_{1}^{4} \frac{1}{u}\, \mathrm{d}u = \Big[\log |u| \Big]_{1}^{4} = \log 4 - \log 1 = \log 4.$$