常见一元函数微分和积分公式

一些基本的微分和积分

$$ \begin{array}{|c|c|c|} \hline f(x)&f^{'}(x)&\int f(x)\,{\rm d}x \\\hline x^n&nx^{n-1}&\frac1{n+1}x^{n+1}\\ a^x&\ln a·a^x&\frac1{\ln a}a^x\\ \ln x&\frac1x&x \ln x- x\\\hline \sin x & \cos x & -\cos x\\ \cos x & -\sin x & \sin x\\\hline \sec x & \sec x \tan x & \ln | \sec x + \tan x | \\ \csc x & -\csc x \cot x & \ln | \csc x - \cot x | \\\hline \tan x & \sec^2x & -\ln | \cos x |\\ \cot x & - \csc^2 x &\ln | \sin x |\\\hline \arcsin x & \frac1{\sqrt{1-x^2}} & x \arcsin x + 2\sqrt{1-x^2}\\ \arctan x & \frac1{1+x^2} & x \arctan x-\frac12ln(1+x^2)\\\hline \end{array} $$

一些常见的微分和积分

$$ \begin{array}{|c|c|c|} \hline f(x) & f^{'}(x) & \int f(x) \,{\rm d}x\\\hline \frac1x&-\frac1{x^2} & \ln x\\\hline \frac1{x^n}&-\frac n{x^{n+1}} & -\frac1{n+1} ·\frac{1}{x^{(n+1)}}\\\hline \sqrt{1-x^2} & \frac{-x}{\sqrt{1-x^2}} & \frac12 (\sqrt{1-x^2} + \arcsin x)\\\hline \sqrt{x^2\pm1}&\frac{±x}{\sqrt{1\pm x^2}} & \frac12(x\sqrt{x^2 \pm 1} \pm \ln|x + \sqrt{x^2 \pm 1}|) \\\hline \frac1{\sqrt{1-x^2}} & x{{(1-x^2)}^{-\frac32}} & \arcsin x\\\hline \frac1{\sqrt{x^2\pm1}}&{\mp x}{{(1\pm x^2)}^{-\frac32}} & \ln |\sqrt{x^2\pm 1} + x| \\\hline \frac1{x^2 + 1} & -2x{(x^2 + 1)}^{-2} & \arctan x\\\hline \frac1{x^2 - 1} & -2x(x^2 - 1)^{-2} & \frac12 \ln|\frac{x-1}{x+1}| \\\hline \end{array} $$

一些积分公式的推导

csc(x)

$$ \begin{aligned} \int \csc x&=\frac1{\sin x}\,{\rm d}x\\ &= \int \frac1{2\sin\frac x2 \cos\frac x2}\,{\rm d}x\\ &= \int \frac1{2\tan\frac x2 \cos^2\frac x2}\,{\rm d}x\\ &= \int \frac{\sec^2\frac x2}{2\tan\frac x2}\,{\rm d}x\\ &= \int \frac1{\tan\frac x2}\,{\rm d}{\tan\frac x2}\\ &= \ln|\tan \frac x2| + C\\ \end{aligned} $$

而

$$ \begin{aligned} \tan\frac x2 = \frac{\sin\frac x2}{\cos\frac x2}= \frac{\sin^2\frac x2}{\sin\frac x2\cos\frac x2}= \frac{1 - \ \cos x}{\sin x}= \csc x - \cot x \end{aligned} $$

从而

$$ \begin{aligned} \int \csc x&=\frac1{\sin x}\,{\rm d}x\\ &= \ln|\tan \frac x2| + C\\ &= \ln|\csc x - \cot x| + C \end{aligned} $$

sec(x)

$$ \begin{aligned} \int \sec x\,{\rm d}x & = \int \csc (x + \frac{\pi}2) \,{\rm d}(x + \frac{\pi}2)\\ & = \ln|\csc (x + \frac{\pi}2) - \cot (x + \frac{\pi}2)| + C\\ & = \ln|\sec x + \tan x| + C \end{aligned} $$

sec³(x)

$$ \begin{aligned} \int \sec^3x\,{\rm d}x&=\int \sec x\,{\rm d}\tan x\\ &=\sec x\,\tan x-\int \tan x\,{\rm d}\sec x\,{\rm d}x\\ &=\ secx\tan x-\int \tan^2x\sec x\,{\rm d}x\\ &=\sec x\tan x-\int (\sec^2x-1)\sec x\,{\rm d}x\\ &=secxtanx-\int sec^3x\,{\rm d}x+\int secx\,{\rm d}x\\ 2\int \sec^3x\,{\rm d}x&=\sec x \tan x+\int \sec x\,{\rm d}x\\ &=\sec x\,\tan x+\ln|\sec x + \tan x| + C\\ \\ \int \sec^3x&=\frac12(\sec x\,\tan x+\ln|\sec x + \tan x|) + C\\ \end{aligned} $$

1/√(a²-x²)

$$ \begin{aligned} \int \frac1{\sqrt{a^2-x^2}}\,{\rm d}x&=\frac1a\int \frac1{\sqrt{1-(\frac xa)^2}}\,{\rm d}x\\ &=\int \frac1{\sqrt{1-(\frac xa)^2}}\,{\rm d}\frac xa\\ \int \frac1{\sqrt{a^2-x^2}}\,{\rm d}x&=\arcsin \frac xa + C \end{aligned} $$

1/√(x²±a²)

$$ \begin{aligned} \int \frac1{\sqrt{x^2 + a^2}}\,{\rm d}x&= \int \frac1{\sec t}\,{\rm d}\tan t&(x=a \tan t, t \in(0,\frac\pi2)) \\ &= \int \sec t\,{\rm d}t\\ &= \ln |\sec t + \tan t| + C\\ &= \ln |\frac{\sqrt{x^2+a^2} + x}a| + C\\ &= \ln |\sqrt{x^2+a^2} + x| + C_1\\ \end{aligned} $$

$$ \begin{aligned} \int \frac1{\sqrt{x^2 - a^2}}\,{\rm d}x&= \int \frac1{\tan t}\,{\rm d}\sec t&(x=a \sec t, t \in(0,\frac\pi2)) \\ &= \int \sec t\,{\rm d}t\\ &= \ln |\sec t + \tan t| + C\\ &= \ln |\frac{\sqrt{x^2-a^2} + x}a| + C\\ &= \ln |\sqrt{x^2-a^2} + x| + C_1\\ \end{aligned} $$

从而

$$ \int \frac1{\sqrt{x^2 \pm a^2}}\,{\rm d}x = \ln |\sqrt{x^2\pm a^2} + x| + C $$

√(a²-x²)

$$ \begin{aligned} \int \sqrt{a^2 - x^2}\,{\rm d}x &= a^2 \int \cos t\,{\rm d}\sin t&(x=a \sin t, t \in[-\frac\pi2,\frac\pi2]) \\ &= a^2 \int \cos^2 t\,{\rm d}t\\ &= \frac14a^2\sin2t +\frac12a^2t + C\\ &= \frac12x\sqrt{a^2-x^2} + \frac {a^2}2 \arcsin\frac xa + C\\ \int \sqrt{a^2 - x^2}\,{\rm d}x &= \frac12 \sqrt{1-x^2} + \frac 12 \arcsin x + C \end{aligned} $$

√(x²±a²)

$$ \begin{aligned} \int \sqrt{x^2 - a^2}\,{\rm d}x&= a^2 \int \tan t\,{\rm d}\sec t&(x=a \sec t,t\in(0,\frac\pi2)) \\ &= a^2 \int \tan^2 t \sec t\,{\rm d}t \\ &= a^2 \int (\sec^2 t -1)\sec t\,{\rm d}t \\ &= a^2 \int \sec^3t\,{\rm d}t - a^2\int \sec t \,{\rm d}t\\ &= a^2(\frac12\sec t \tan t - \frac12\ln| \tan t + \sec t|) + C\\ &= \frac12a^2(\frac{x\sqrt{x^2 - a^2}}{a^2} - \ln|\frac {x + \sqrt{x^2 - a^2}}a| + C\\ &= \frac12(x\sqrt{x^2-a^2} - a^2\ln|x+\sqrt{x^2-a^2}|) + C_1 \end{aligned} $$

$$ \begin{aligned} \int \sqrt{x^2 + a^2}\,{\rm d}x&= a^2 \int \sec t\,{\rm d}\tan t&(x=a \tan t,t\in(0,\frac\pi2)) \\ &= a^2 \int \sec^3 t \,{\rm d}t \\ &= a^2(\frac12\sec t \tan t + \frac12\ln| \tan t + \sec t|) + C\\ &= \frac12a^2(x\sqrt{x^2 + a^2} + \ln|\frac {x + \sqrt{x^2 + a^2}}a|) + C\\ &= \frac12a^2(\frac{x\sqrt{x^2 + a^2}}{a^2} + \ln|\frac {x + \sqrt{x^2 + a^2}}a| + C\\ &= \frac12(x\sqrt{x^2+a^2} + a^2\ln|x+\sqrt{x^2-a^2}|) + C_1 \end{aligned} $$

从而

$$ \begin{aligned} \int \sqrt{x^2 \pm a^2}\,{\rm d}x &= \frac12(x\sqrt{x^2 \pm a^2} \pm a^2\ln|\frac {x + \sqrt{x^2 \pm a^2}}a|)+C \\ \int \sqrt{x^2 \pm 1}\,{\rm d}x &= \frac12(x\sqrt{x^2 \pm 1} \pm \ln|x + \sqrt{x^2 \pm 1}|) + C \end{aligned} $$

1/(x² - a²)

$$ \begin{aligned} \int \frac1{x^2 - a^2}\,{\rm d}x&= \int \frac1{(x + a)(x - a)}\,{\rm d}x\\ &= \frac1{2a} \int (\frac1{x - a} - \frac1{x + a}) \,{\rm d}x\\ &= \frac1{2a} \ln|\frac{x-a}{x+a}| + C \end{aligned} $$

1/(x² + a²)

$$ \begin{aligned} \int \frac1{x^2 + a^2}\,{\rm d}x&= \frac 1{a^2} \int \frac1{(\frac xa)^2+1}\,{\rm d}x\\ &= \frac1a \int \frac1{(\frac xa)^2+1}\,{\rm d}{\frac xa}\\ &= \frac1a \arctan{\frac xa} + C \end{aligned} $$