基本积分表

exordium

基本积分表是求不定积分的基础，求解不定积分，往往是利用各种方法将其变形、分解，最终表述为基本积分表的形式，从而得解。

下面列举数个基本积分公式的推导过程，其余可在此处之内回推： 常见函数导数

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First

$$\begin{align} \text { 前知识1: } y=\arctan (x) \Leftrightarrow \tan (y)=x \\ \text { 前知识2: }(\tan \alpha)^{\prime}=\sec ^{2} \alpha \\ \text { 前知识3: } \sec ^{2} \alpha=1+\tan ^{2} \alpha \\ \\ (\arctan x)^{\prime}=\text { ? } \\ \\ y'=[\arctan (x)]' \Leftrightarrow {[\tan (y)]^{\prime}=(x)^{\prime}} \\ \\ \because y=\arctan (x) \\ \\ {因此 [\tan (y)] 为复合函数} \\ \\ 链式法则: [\tan (y)]^{\prime} \cdot y^{\prime}=(x)^{\prime} \\ \\ \sec ^{2}y \cdot y^{\prime}=1 \\ \\ \because \text { 前知识3 } \\ \\ \Rightarrow\left(1+\tan ^{2} y\right) \cdot y^{\prime}=1 \\ \\ \because \tan (y)=x \\ \therefore \tan ^{2} y=x^{2} \\ \\ (1 + x^2) \cdot y' = 1 \\ \\ \Rightarrow y' = \frac{1}{1 + x^2} \\ \\ \therefore \frac{d}{dx} [\arctan(x)] = \frac{1}{1 + x^2} \\ \\ \therefore \int \frac{1}{1+x^{2}} dx=\arctan x +C \end{align}$$

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Second

$$\begin{eqnarray} \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C \\ \mu 为常数, 但\mu \ne 0 \\ \\ 推导过程如下: \\ (\frac{1}{\mu+1} x^{\mu+1})' =\frac{(x^{\mu+1})' \cdot \mu +1 - x^{\mu+1} \cdot (\mu+1)'}{(\mu+1)^{2}} \\ \\ \because (\mu+1)'=0, \enspace (x^{\mu+1})'=(\mu+1) x^{\mu+1-1} \\ \\ \Rightarrow \frac{(\mu+1) x^{\mu} \cdot \mu +1 - x^{\mu+1} \cdot 0}{(\mu+1)^{2}} \\ \\ \frac{((\mu+1) x^{\mu} \cdot \mu +1}{(\mu+1)^{2}}=x^{\mu} \\ \\ (\frac{1}{\mu+1} x^{\mu+1})'=x^{\mu} \\ \\ \therefore \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C \end{eqnarray}$$

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Third

$$\begin{eqnarray} \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C, \enspace (a为常数,a>0,a \ne 1) \\ \\ 推导过程如下: \\ 已知: \enspace (a^{x})'=a^{x}\ln{a}, \enspace (\ln{a})'=0 \\ \\ \frac{(a^{x})'}{(\ln{a})'}=\frac{(a^{x})'\ln{a}-a^{x}(\ln{a})'}{\ln^{2}{a}} \\ \\ \frac{a^{x}\ln{a} \cdot \ln{a}-a^{x}\cdot 0}{\ln^{2}{a}} \\ \\ \Rightarrow \frac{a^{x}\ln{a} \cdot \ln{a}}{\ln^{2}{a}}=a^{x} \\ \\ \therefore (\frac{a^{x}}{\ln{a}})'=a^{x} \\ \\ \therefore \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C , \enspace (a>0,a \ne 1) \end{eqnarray}$$

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Table

• $\int \frac{1}{1+x^{2}} dx = \arctan x+C$
  
  
• $\int \frac{1}{\sqrt[]{1-x^{2}}} dx = \arcsin x+C$
  
  
• $\int \sec x dx=\ln |\sec x+\tan x| + C$
  推导过程 - Sample 0
  
  

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