> [!important] > > 请**一定牢记**接下来的公式。因为考试时，所有的积分计算都将化为下方的公式。   ## 最重要的几个公式 - $\int a^{x}\mathrm{d}x = \frac{a^{x}}{\ln a} +C ~ (a > 0, a \ne 1)$ ### 原函数为三角函数 - $\int \tan x \mathrm{d}x = -\ln | \cos x| +C$ > [!info] > > $\tan x$ 积分证明推导： > > $\begin{align*} & \int \tan x \mathrm{d}x\\ =& \int \frac{\sin x}{\cos x} \mathrm{d}x\\ =& -\int \frac{1}{\cos x} (\cos x)' \mathrm{d}x \end{align*}$ > > 令 $\cos x = U$ > > $\begin{align*} =& -\int \frac{1}{U} \mathrm{d}U \\ =& -\ln |U| + C \\ =& -\ln |\cos x| + C \end{align*}$ - $\int \frac{\mathrm{d}x}{\cos x} = \int \sec x \mathrm{d}x = \ln |\sec x + \tan x| + C$ - $\int \frac{\mathrm{d}x}{\sin x} = \int \csc x \mathrm{d}x = \ln |\csc x - \cot x| + C$ --- ### 原函数为 $\frac{1}{m^{2} \pm n^{2}}$ #### 不含根号 - $\int \frac{1}{a^{2}+x^{2}} \mathrm{d}x = \frac{1}{a}\arctan \frac{x}{a}+C$ - $\int \frac{1}{x^{2}-a^{2}} \mathrm{d}x = \frac{1}{2a}\ln \left| \frac{x-a}{x+a} \right| + C$ - $\int \frac{1}{a^{2}-x^{2}} \mathrm{d}x = \frac{1}{2a}\ln \left| \frac{x+a}{x-a} \right| + C$ #### 含根号 - $\int \frac{1}{\sqrt{x^{2}-a^{2}}} \mathrm{d}x = \ln \left | x + \sqrt{x^{2}-a^{2}} \right | + C$ - $\int \frac{1}{\sqrt{a^{2}-x^{2}}} \mathrm{d}x = \arcsin \frac{x}{a}+C$ - $\int \sqrt{a^{2}-x^{2}} ~\mathrm{d}x = \frac{a^{2}}{2}\arcsin \frac{x}{a} + \frac{x}{2} \sqrt{a^{2}-x^{2}} + C$