微积分常用公式表
微分、導數和積分公式對照表
| | | 冪函數 | ||
| 1 | d(xμ)=μxμ?1dx\mathbfozvdkddzhkzd(x^\mu)=\mu x^{\mu-1}\ \mathbfozvdkddzhkzdxd(xμ)=μxμ?1?dx | (xμ)′=μxμ?1(x^\mu)'=\mu x^{\mu-1}(xμ)′=μxμ?1 | ∫xμdx=xμ+1μ+1+C\int x^\mu \mathbfozvdkddzhkzdx=\cfrac{x^{\mu+1}}{\mu+1}+C∫xμdx=μ+1xμ+1?+C |
| | | 指數函數 | ||
| 2 | d(ax)=axln?adx\mathbfozvdkddzhkzd(a^x)=a^x \ln a\ \mathbfozvdkddzhkzdxd(ax)=axlna?dx | (ax)′=axln?a(a^x)'=a^x \ln a(ax)′=axlna | ∫axdx=axln?a+C\int a^x\mathbfozvdkddzhkzdx=\cfrac{a^x}{\ln a}+C∫axdx=lnaax?+C |
| 3 | d(ex)=exdx\mathbfozvdkddzhkzd(e^x)=e^x\ \mathbfozvdkddzhkzdxd(ex)=ex?dx | (ex)′=ex(e^x)'=e^x(ex)′=ex | ∫exdx=ex+C\int e^x \mathbfozvdkddzhkzdx=e^x+C∫exdx=ex+C |
| | | 對數函數 | ||
| 4 | d(log?ax)=1xln?adx\mathbfozvdkddzhkzd(\log_ax)=\cfrac{1}{x\ln a}\mathbfozvdkddzhkzdxd(loga?x)=xlna1?dx | (log?ax)′=1xln?a(\log_ax)'=\cfrac{1}{x\ln a}(loga?x)′=xlna1? | |
| 5 | d(ln?x)=1xdx\mathbfozvdkddzhkzd(\ln x)=\cfrac{1}{x}\mathbfozvdkddzhkzdxd(lnx)=x1?dx | (ln?x)′=1x(\ln x)'=\cfrac{1}{x}(lnx)′=x1? | ∫1xdx=ln?∣x∣+C\int \cfrac{1}{x} \mathbfozvdkddzhkzdx=\ln \vert x \vert+C∫x1?dx=ln∣x∣+C |
| | | 三角函數 | ||
| 6 | d(sin?x)=cos?xdx\mathbfozvdkddzhkzd(\sin x)=\cos x\ \mathbfozvdkddzhkzdxd(sinx)=cosx?dx | (sin?x)′=cos?x(\sin x)'=\cos x(sinx)′=cosx | ∫cos?xdx=sin?x+C\int \cos x\mathbfozvdkddzhkzdx=\sin x+C∫cosxdx=sinx+C |
| 7 | d(cos?x)=?sin?xdx\mathbfozvdkddzhkzd(\cos x)=-\sin x\ \mathbfozvdkddzhkzdxd(cosx)=?sinx?dx | (cos?x)′=?sin?x(\cos x)'=-\sin x(cosx)′=?sinx | ∫sin?xdx=?cos?x+C\int \sin x \mathbfozvdkddzhkzdx=-\cos x+C∫sinxdx=?cosx+C |
| 8 | d(tan?x)=sec?2xdx\mathbfozvdkddzhkzd(\tan x)=\sec^2x\ \mathbfozvdkddzhkzdxd(tanx)=sec2x?dx | (tan?x)′=sec?2x(\tan x)'=\sec^2x(tanx)′=sec2x | ∫sec?2xdx=∫1cos?2x=tan?x+C\int \sec^2x\mathbfozvdkddzhkzdx=\int \cfrac{1}{\cos^2x}=\tan x+C∫sec2xdx=∫cos2x1?=tanx+C |
| | | |||
| 9 | d(cot?x)=?csc?2xdx\mathbfozvdkddzhkzd(\cot x)=-\csc^2x\ \mathbfozvdkddzhkzdxd(cotx)=?csc2x?dx | (cot?x)′=?csc?2x(\cot x)'=-\csc^2x(cotx)′=?csc2x | ∫csc?2xdx=∫1sin?2x=?cot?x+C\int \csc^2x\mathbfozvdkddzhkzdx=\int \cfrac{1}{\sin^2x}=-\cot x+C∫csc2xdx=∫sin2x1?=?cotx+C |
| 10 | d(sec?x)=sec?xtan?xdx\mathbfozvdkddzhkzd(\sec x)=\sec x\ \tan x\ \mathbfozvdkddzhkzdxd(secx)=secx?tanx?dx | (sec?x)′=sec?xtan?x(\sec x)'=\sec x\tan x(secx)′=secxtanx | ∫sec?xtan?xdx=sec?x+C\int \sec x \tan x\mathbfozvdkddzhkzdx=\sec x+C∫secxtanxdx=secx+C |
| 11 | d(csc?x)=?csc?xcot?xdx\mathbfozvdkddzhkzd(\csc x)=-\csc x\cot x\ \mathbfozvdkddzhkzdxd(cscx)=?cscxcotx?dx | (csc?x)′=?csc?xcot?x(\csc x)'=-\csc x\cot x(cscx)′=?cscxcotx | ∫csc?xcot?xdx=?csc?x+C\int \csc x \cot x \mathbfozvdkddzhkzdx=-\csc x+C∫cscxcotxdx=?cscx+C |
| | | 反三角函數 | ||
| 12 | d(arcsin?x)=11?x2dx\mathbfozvdkddzhkzd(\arcsin x)=\cfrac{1}{\sqrt{1-x^2}}\mathbfozvdkddzhkzdxd(arcsinx)=1?x2?1?dx | (arcsin?x)′=11?x2(\arcsin x)'=\cfrac{1}{\sqrt{1-x^2}}(arcsinx)′=1?x2?1? | ∫11?x2=arcsin?x+C\int \cfrac{1}{\sqrt{1-x^2}}=\arcsin x+C∫1?x2?1?=arcsinx+C |
| 13 | d(arccos?x)=?11?x2dx\mathbfozvdkddzhkzd(\arccos x)=-\cfrac{1}{\sqrt{1-x^2}}\mathbfozvdkddzhkzdxd(arccosx)=?1?x2?1?dx | (arccos?x)′=?11?x2(\arccos x)'=-\cfrac{1}{\sqrt{1-x^2}}(arccosx)′=?1?x2?1? | |
| 14 | d(arctan?x)=11+x2dx\mathbfozvdkddzhkzd(\arctan x)=\cfrac{1}{1+x^2}\mathbfozvdkddzhkzdxd(arctanx)=1+x21?dx | (arctan?x)′=11+x2(\arctan x)'=\cfrac{1}{1+x^2}(arctanx)′=1+x21? | ∫11+x2=arctan?x+C\int \cfrac{1}{1+x^2}=\arctan x+C∫1+x21?=arctanx+C |
| 15 | d(arccotx)=?11+x2dx\mathbfozvdkddzhkzd(arccot\ x)=-\cfrac{1}{1+x^2}\mathbfozvdkddzhkzdxd(arccot?x)=?1+x21?dx | (arccotx)′=?11+x2(arccot x)'=-\cfrac{1}{1+x^2}(arccotx)′=?1+x21? |
積分續表
∫kdx=kx+C,(k、C是常數)\int k \mathbfozvdkddzhkzdx=kx+C,(k、C是常數)∫kdx=kx+C,(k、C是常數)
∫tan?xdx=?ln?∣cos?x∣+C\int \tan x\mathbfozvdkddzhkzdx=-\ln |\cos x|+C∫tanxdx=?ln∣cosx∣+C
∫cot?xdx=ln?∣sin?x∣+C\int \cot x\mathbfozvdkddzhkzdx=\ln |\sin x|+C∫cotxdx=ln∣sinx∣+C
∫sec?xdx=ln?∣sec?x+tan?x∣+C\int \sec x\mathbfozvdkddzhkzdx=\ln|\sec x+\tan x|+C∫secxdx=ln∣secx+tanx∣+C
∫csc?xdx=ln?∣csc?x?cot?x∣+C\int \csc x\mathbfozvdkddzhkzdx=\ln|\csc x-\cot x|+C∫cscxdx=ln∣cscx?cotx∣+C
∫1a2+x2dx=1aarctan?xa+C\int \cfrac{1}{a^2+x^2}\mathbfozvdkddzhkzdx=\cfrac{1}{a}\arctan \cfrac{x}{a}+C∫a2+x21?dx=a1?arctanax?+C
∫1x2?a2dx=12aln?∣x?ax+a∣+C\int \cfrac{1}{x^2-a^2}\mathbfozvdkddzhkzdx=\cfrac{1}{2a}\ln \vert \dfrac{x-a}{x+a} \vert + C∫x2?a21?dx=2a1?ln∣x+ax?a?∣+C
∫1a2?x2dx=arcsin?xa+C\int \cfrac{1}{\sqrt{a^2-x^2}}\mathbfozvdkddzhkzdx=\arcsin \cfrac{x}{a}+C∫a2?x2?1?dx=arcsinax?+C
∫1x2+a2dx=ln?∣x+x2+a2∣+C\int \cfrac{1}{\sqrt{x^2+a^2}}\mathbfozvdkddzhkzdx=\ln |x+\sqrt{x^2+a^2}|+C∫x2+a2?1?dx=ln∣x+x2+a2?∣+C
∫1x2?a2dx=ln?∣x+x2?a2∣+C\int \cfrac{1}{\sqrt{x^2-a^2}}\mathbfozvdkddzhkzdx=\ln|x+\sqrt{x^2-a^2}|+C∫x2?a2?1?dx=ln∣x+x2?a2?∣+C
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