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
First
\[\begin{align}
\text { 前知识1: } y=\arctan (x) \Leftrightarrow \tan (y)=x
\\
\text { 前知识2: }(\tan \alpha)^{\prime}=\sec ^{2} \alpha
\\
\text { 前知识3: } \sec ^{2} \alpha=1+\tan ^{2} \alpha
\\ \\
(\arctan x)^{\prime}=\text { ? }
\\ \\
y'=[\arctan (x)]' \Leftrightarrow {[\tan (y)]^{\prime}=(x)^{\prime}}
\\ \\
\because y=\arctan (x)
\\ \\
{因此 [\tan (y)] 为复合函数}
\\ \\
链式法则: [\tan (y)]^{\prime} \cdot y^{\prime}=(x)^{\prime}
\\ \\
\sec ^{2}y \cdot y^{\prime}=1
\\ \\
\because \text { 前知识3 }
\\ \\
\Rightarrow\left(1+\tan ^{2} y\right) \cdot y^{\prime}=1
\\ \\
\because \tan (y)=x
\\
\therefore \tan ^{2} y=x^{2}
\\ \\
(1 + x^2) \cdot y' = 1
\\ \\
\Rightarrow y' = \frac{1}{1 + x^2}
\\ \\
\therefore \frac{d}{dx} [\arctan(x)] = \frac{1}{1 + x^2}
\\ \\
\therefore \int \frac{1}{1+x^{2}} dx=\arctan x +C
\end{align}
\]
Second
\[\begin{eqnarray}
\int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C
\\
\mu 为常数, 但\mu \ne 0
\\ \\
推导过程如下: \\
(\frac{1}{\mu+1} x^{\mu+1})'
=\frac{(x^{\mu+1})' \cdot \mu +1 - x^{\mu+1} \cdot (\mu+1)'}{(\mu+1)^{2}}
\\ \\
\because (\mu+1)'=0, \enspace (x^{\mu+1})'=(\mu+1) x^{\mu+1-1}
\\ \\
\Rightarrow \frac{(\mu+1) x^{\mu} \cdot \mu +1 - x^{\mu+1} \cdot 0}{(\mu+1)^{2}}
\\ \\
\frac{((\mu+1) x^{\mu} \cdot \mu +1}{(\mu+1)^{2}}=x^{\mu}
\\ \\
(\frac{1}{\mu+1} x^{\mu+1})'=x^{\mu}
\\ \\
\therefore \int x^{\mu}dx=\frac{1}{\mu+1} x^{\mu+1}+C
\end{eqnarray}
\]
Third
\[
\begin{eqnarray}
\int a^{x}dx=\frac{a^{x}}{\ln{a}}+C, \enspace (a为常数,a>0,a \ne 1)
\\ \\
推导过程如下: \\
已知: \enspace (a^{x})'=a^{x}\ln{a}, \enspace
(\ln{a})'=0
\\ \\
\frac{(a^{x})'}{(\ln{a})'}=\frac{(a^{x})'\ln{a}-a^{x}(\ln{a})'}{\ln^{2}{a}}
\\ \\
\frac{a^{x}\ln{a} \cdot \ln{a}-a^{x}\cdot 0}{\ln^{2}{a}}
\\ \\
\Rightarrow \frac{a^{x}\ln{a} \cdot \ln{a}}{\ln^{2}{a}}=a^{x}
\\ \\
\therefore (\frac{a^{x}}{\ln{a}})'=a^{x}
\\ \\
\therefore \int a^{x}dx=\frac{a^{x}}{\ln{a}}+C
, \enspace (a>0,a \ne 1)
\end{eqnarray}
\]
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
