题

$$\begin{array}{rl}& \\ & \frac{1}{2}\int e^x{\left(\tan \frac{x}{2}+1\right)}^{2}dx\\ & =\frac{1}{2}{e}^x{\left(\tan \frac{x}{2}+1\right)}^2-\int e^x\left(\tan \frac{x}{2}+1\right)d\tan \frac{x}{2}\\ & \\ & =\frac{1}{2}{e}^x{\left(\tan \frac{x}{2}+1\right)}^2-\int e^x\tan \end{array}$$

为什么

$$\underset{x\to 0}{\lim }{e}^x-1-x=\underset{x\to 0}{\lim }k{x}^2$$

k可以取任意实数，而

$$\underset{x\to 0}{\lim }\frac{e^x-1-x}{x^2}=k$$

k只能取 $\frac{1}{2}$

$$\int x\ln (1+x)dx=\frac{1}{2}{x}^2\ln (1+x)-\frac{1}{4}{x}^2+\frac{x}{2}-\frac{1}{2}\ln (1+x)+C$$ $${\int }_{}^{}\ln (1+x^2)dx=x\ln (1+x^2)-2x+2\arctan x+C$$ $$\underset{r\to 0}{\lim }\frac{-6r^2\cos \theta }{6r^2\sin \theta -4{\sin }^3\theta }$$ $$\cos (\arctan a)=?$$ $$\sqrt{e^x}=(e^x{)}^{1/2}=e^{x/2}$$ $$\begin{array}{r}a\uparrow b=a^b\\ a\uparrow \uparrow b=a\uparrow a\uparrow a\cdots a=a^{a^{a^a}}(b个a)\end{array}$$