常用初等函数微积分

不定积分需要 $+C$,下面都省略了。验证两个不定积分结果是否相等的方法是求导。

导数积分
$(x^n)'=nx^{n-1}$$\int x^n{\rm d}x=\dfrac{x^{n+1}}{n+1}$
$(e^x)'=e^x$$\int e^x{\rm d}x=e^x$
$(a^x)'=a^x\ln a$$\int a^x{\rm d}x=\dfrac{a^x}{\ln a}$
$(\ln x)'=\dfrac1x$$\int\ln x{\rm d}x=x\ln x-x$
$(\log_ax)'=\dfrac1{x\ln a}$$\int\log_ax{\rm d}x=\dfrac{1}{\ln a}(x\ln x-x)$
$\sin'x=\cos x$$\int\sin x{\rm d}x=-\cos x$
$\cos'x=-\sin x$$\int\cos x{\rm d}x=\sin x$
$\tan'x=\sec^2x$$\int\tan x{\rm d}x=-\ln\lvert\cos x\rvert$
$\cot'x=-\csc^2x$$\int\cot x{\rm d}x=\ln\lvert\sin x\rvert$
$\sec'x=\sec x\tan x$$\int\sec x{\rm d}x=\ln\lvert\sec x+\tan x\rvert$
$\csc'x=-\csc x\cot x$$\int\csc x{\rm d}x=\ln\lvert\csc x-\cot x\rvert$
$\arcsin'x=\dfrac{1}{\sqrt{1-x^2}}$
$\arctan'x=\dfrac{1}{1+x^2}$
$\int\dfrac{1}{x^2-a^2}{\rm d}x=\dfrac{1}{2a}\ln\lvert\dfrac{x-a}{x+a}\rvert$
$\int\dfrac{1}{x^2+a^2}{\rm d}x=\dfrac{1}{a}\arctan\dfrac{x}{a}$
$\int\dfrac{1}{\sqrt{a^2-x^2}}{\rm d}x=\arcsin\dfrac{x}{a}$
$\int\dfrac{1}{\sqrt{x^2+a^2}}{\rm d}x=\ln(x+\sqrt{x^2+a^2})$
$\int\dfrac{1}{\sqrt{x^2-a^2}}{\rm d}x=\ln\lvert x+\sqrt{x^2-a^2}\rvert$
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