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4.1 不定积分概念与性质

§4.1\S 4.1 不定积分概念与性质#

一、原函数与不定积分#

  • 定义:f(x)f(x)F(x)F(x)II 上有定义, F(x)=f(x)F'(x)=f(x),则称 F(x)F(x) 为原函数。d(x3)=3x2dx\mathrm d(x^3)=3x^2\mathrm dx,则称 x3x^33x23x^2一个原函数
  1. 存在性定理:若初等函数在定义区间连续,则在该定义区间上存在原函数 F(x)F(x)
  2. f(x)f(x) 的全体原函数构成的集合,称为不定积分,记作
积分号f(x)被积函数dx积分变量=F(x)+C积分常数\underset{积分号}{\int} \underset{被积函数}{f(x)}\mathrm d\underset{积分变量}{x}=F(x)+\underset{积分常数}{C}
  1. 初等函数的原函数不一定是初等函数

二、基本积分表#

请熟练记住它们,不要忘记积分常数!

  1. kdx=kx+C\displaystyle \int k\mathrm dx=kx+Ckk 为常数
  2. xαdx=1α+1xα+1+C\displaystyle \int x^\alpha \mathrm dx=\dfrac{1}{\alpha+1}x^{\alpha+1}+Cα1x>0\alpha \ne -1\quad x\gt 0
  3. 1xdx=lnx+C\displaystyle \int \dfrac{1}{x}\mathrm dx=\ln |x|+C
  4. 11+x2dx=arctanx+C\displaystyle \int \dfrac{1}{1+x^2}\mathrm dx=\arctan x+C
  5. 11x2dx=arcsinx+C\displaystyle \int \dfrac{1}{\sqrt{1-x^2}}\mathrm dx=\arcsin x+C
  6. cosxdx=sinx+C\displaystyle \int \cos x\mathrm dx=\sin x+C
  7. sinxdx=cosx+C\displaystyle \int \sin x\mathrm dx=-\cos x+C
  8. sec2xdx=tanx+C\displaystyle \int \sec^2 x\mathrm dx=\tan x+C
  9. csc2xdx=cotx+C\displaystyle \int \csc^2 x\mathrm dx=-\cot x+C
  10. secxtanxdx=secx+C\displaystyle \int \sec x\tan x\mathrm dx=\sec x+C
  11. cscxcotxdx=cscx+C\displaystyle \int \csc x\cot x\mathrm dx=-\csc x+C
  12. exdx=ex+C\displaystyle \int e^x\mathrm dx=e^x+C
  13. axdx=axlna+C\displaystyle \int a^x\mathrm dx=\dfrac{a^x}{\ln a}+Ca>0a \gt 0a1a\ne 1

三、性质(线性)#

定理:f(x)f(x)g(x)g(x)II 上均存在原函数,λ,μ\lambda,\mu为任意实数,则

[λf(x)+μg(x)]dx=λf(x)dx+μg(x)dx\int [\lambda f(x)+\mu g(x)]\mathrm dx=\lambda \int f(x)\mathrm dx+\mu \int g(x)\mathrm dx

例题#

(a0+a1x+a2x2++anxn)dx=a01dx+a1xd+anxndx=a0(x+C1)+a1(x22+C2)++an(xn+1n+1+Cn)=a0x+a1x22++anxn+1n+1+C\begin{align*} \int (a_0+a_1x+a_2x^2+\cdots+a_nx^n)\mathrm dx&=a_0\int 1\mathrm dx+a_1\int x\mathrm d\cdots+a_n\int x_n\mathrm dx\\ &=a_0(x+C_1)+a_1(\dfrac{x^2}{2}+C_2)+\cdots+a_n(\dfrac{x^{n+1}}{n+1}+C_n)\\ &=a_0x+a_1\dfrac{x^2}{2}+\cdots+a_n\dfrac{x^{n+1}}{n+1}+C \end{align*}
(bnxn+bn1xn+1++b1x1+b0)dx=bnxn+1n+1+bn1xn+2n+2++b2x12+1+b1lnx+b0x+C\begin{align*} \int (b_nx^{-n}+b_{n-1}x^{-n+1}+\cdots+b_1x^{-1}+b_0)\mathrm dx&=b_n\dfrac{x^{-n+1}}{-n+1}+b_{n-1}\dfrac{x^{-n+2}}{-n+2}+\cdots+b_2\dfrac{x^{-1}}{-2+1}+b_1\ln |x|+b_0x+C \end{align*}
(5x+5x)2dx=(52x+52x+2)dx=52x2ln5+52x2ln5+2x+C\begin{align*} \int (5^x+5^{-x})^2\mathrm dx&=\int (5^{2x}+5^{-2x}+2)\mathrm dx\\ &=\dfrac{5^{2x}}{2\ln 5}+\dfrac{5^{-2x}}{-2\ln 5}+2x+C \end{align*}
1(cosxsinx)2dx=cos2x+sin2xcos2xsin2xdx=(1sin2x+1cos2x)dx=cotx+tanx+C\begin{align*} \int \dfrac{1}{(\cos x\sin x)^2}\mathrm dx&=\int \dfrac{\cos^2 x+\sin^2 x}{\cos^2x\sin^2x}\mathrm dx\\ &=\int (\dfrac{1}{\sin^2x}+\dfrac{1}{\cos^2x})\mathrm dx\\ &=-\cot x+\tan x+C \end{align*}
tan2xdx=(sec2x1)dx=tanxx+C\begin{align*} \int \tan^2x\mathrm dx&=\int (\sec^2 x-1)\mathrm dx\\ &=\tan x-x+C \end{align*}
(cosxsinx)dx=(12sin2x)dx=14cos2x+C\begin{align*} \int (\cos x\sin x)\mathrm dx&=\int (\dfrac{1}{2}\sin 2x)\mathrm dx\\ &=-\dfrac{1}{4}\cos 2x+C \end{align*}
  1. [积化和差]
(cosxsin3x)dx=12(sin4x+sin2x)dx=12(14cos4x12cos2x)+C=18cos4x14cos2x+C\begin{align*} \int (\cos x\sin 3x)\mathrm dx&=\dfrac{1}{2}\int (\sin 4x+\sin 2x)\mathrm dx\\ &=\dfrac{1}{2}(-\dfrac{1}{4}\cos 4x-\dfrac{1}{2}\cos 2x)+C\\ &=-\dfrac{1}{8}\cos 4x-\dfrac{1}{4}\cos 2x+C \end{align*}
x41+x2dx=x41+1x2+1=(x21)dx+11+x2dx=13x3x+arctanx+C\begin{align*} \int \dfrac{x^4}{1+x^2}\mathrm dx&=\int \dfrac{x^4-1+1}{x^2+1}\\ &=\int (x^2-1)\mathrm dx+\int \dfrac{1}{1+x^2}\mathrm dx\\ &=\dfrac{1}{3}x^3-x+\arctan x+C \end{align*}
e3x+1ex+1=(ex+1)(e2xex+1)ex+1=(e2xex+1)=12e2xex+x+C\begin{align*} \int \dfrac{e^{3x}+1}{e^x+1}&=\int \dfrac{(e^x+1)(e^{2x}-e^x+1)}{e^x+1}\\ &=\int (e^{2x}-e^x+1)\\ &=\dfrac{1}{2}e^{2x}-e^x+x+C \end{align*}
  1. 曲线在 xx 点的切线斜率为 x4\dfrac{x}{4},且过 (2,52)(2,\dfrac{5}{2}),求曲线方程

解:y=f(x)y=f(x)f(2)=52f(2)=\dfrac{5}{2}f(x)=x4f'(x)=\dfrac{x}{4}

f(x)=x4dx=14x22+Cf(x)=\int \dfrac{x}{4}\mathrm dx=\dfrac{1}{4}\cdot \dfrac{x^2}{2}+C

代入 f(2)=52f(2)=\dfrac{5}{2} 得到 C=2C=2

y=18x2+2\therefore y=\dfrac{1}{8}x^2+2 即为所求

4.1 不定积分概念与性质
https://gitee.com/jason_ren/advanced-math-note
作者
Jason Ren
发布于
2025-12-03
许可协议
CC BY-SA 4.0

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