치환적분법

• Solution 1: If we let $u=\sqrt{\log x}$, then $\mathrm{d}u=\dfrac{1}{2x\sqrt{\log x}}\ \mathrm{d}x$, and:

$$\begin{aligned} \int \dfrac{e^{ \sqrt{\log x}}}{ 2x\sqrt{\log x}}\ \mathrm{d}x&=\int e^{{red}u}\ \mathrm{d}u=e^u+C=e^{\sqrt{\log x}}+C \end{aligned}$$

• Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it's not unnatural to think that $u=\log x$, $\mathrm{d}u=\dfrac{1}{x}\ \mathrm{d}x$ would be a good choice. In that case:

$$\begin{aligned} \int \dfrac{e^{\sqrt{ \log x}}}{2 x\sqrt{ \log x}}\ \mathrm{d}x&= \int \frac{e^{\sqrt{{red}u}}}{2\sqrt{{red}u}} \mathrm{d}u \\ \text{Now, we should be able to see that $ w=\sqrt{u}$, $ \mathrm{d}w = \dfrac{1}{2\sqrt{u}}\ \mathrm{d}u$ is a good choice:} \\ \int\frac{e^{ \sqrt{u}}}{ 2\sqrt{u}}\ \mathrm{d}u &=\int e^{{orange}w}\ \mathrm{d}w\\ &=e^{\sqrt{u}}+C\\ &=e^{\sqrt{\log x}}+C \end{aligned}$$