数学公式

等价无穷小

注释：以下 $x$ 均当作成一个整体，不仅仅单指 $x$ ；$x \to 0$ 时以下成立
$$sinx \sim x \qquad tanx \sim x \qquad arcsinx \sim x \qquad arctanx \sim x \qquad 1-cosx \sim \frac{1}{2}x^2$$

$$ ln(1+x) \sim x \qquad (1+x)^\alpha-1 \sim \alpha x \qquad e^x - 1 \sim x \qquad a^x-1 \sim xlna \quad(a>0)$$

$$log_a(1+x) \sim \frac{x}{lna} \qquad x-ln(1+x)\sim\frac{1}{2}x^2 \qquad ln(x+\sqrt{1+x^2}) \sim x \qquad x-sinx \sim \frac{1}{6}x^3$$

$$tanx-x \sim \frac{1}{3}x^3 \qquad arcsinx -x \sim \frac{1}{6}x^3 \qquad x-arctanx \sim \frac{1}{3}x^3 \qquad tanx -sinx \sim \frac{1}{2}x^3$$

$$(1 + \beta x)^\alpha-1 \sim \alpha\beta x \qquad \sqrt[n]{1+x}-1 \sim \frac{1}{n}x$$

以上等价无穷小替换实质均为：保留到第一阶的在 $x=0$ 处的泰勒展开式（麦克劳林展开）
$$f(x)=f(0)+f'(0)\frac{x}{1!}+f''(0)\frac{x^2}{2!}+ \cdots +f^n\frac{x^n}{n!}$$

对数恒等式：

$$f(x)^{g(x)}=e^{lnf(x)^{g(x)}}=e^{g(x)lnf(x)}$$

重要极限：

$$ \lim_{x \to 0 }\frac{sinx}{x} = 1 \qquad \lim_{x \to \infty}(1+\frac{1}{x})^x = e $$

$$\lim_{x \to \infty}(1+\frac{1}{x})^x = e \qquad \lim_{x \to 0}(1+x)^{\frac{1}{x}}=e$$

$$\lim_{x \to \infty}(1-\frac{1}{x})^x=\frac{1}{e} \qquad \lim_{x \to 0}(1-\frac{1}{x})^{\frac{1}{x}}=\frac{1}{e}$$

$$\lim_{n \to \infty}\sqrt[n]{n} = 1$$

不定积分

$$\int{kdx}=kx+C \qquad \int{x^adx}=\frac{x^{a+1}}{a+1}+C \qquad \int\frac{1}{x}dx=ln|x|+C$$
$$\int\frac{1}{1+x^2}dx=arctanx+C \qquad \int\frac{1}{\sqrt{1+x^2}}dx=arcsinx+C \qquad \int{a^x}dx=\frac{a^x}{lna}+C$$
$$\int coxdx = sinx + C \qquad \int sinxdx=-cosx+C $$
$$\int \frac{1}{sin^2x}dx = csc^2dx = -cotx+C \qquad \int \frac{1}{cos^2x}dx = \int sec^2xdx = tanx + C $$
$$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}acrtan(\frac{x}{a})+C \qquad \int e^xdx = e^x + C$$
$$\int \frac{1}{\sqrt{a^2-x^2}}dx=arcsin(\frac{x}{a})+C \qquad \int tanxdx=-ln|cosx|+C$$
$$\int arctanxdx = \frac{1}{2}ln|1+x^2|+C \qquad \int$$