Section01_定义与方法

  1. 原函数 - f(x),F(x)f(x),F(x) 若 F′(x)=f(x)F'(x) = f(x)，称 F(x)F(x) 为 f(x)f(x) 的原函数

Notes

  1. 若 f(x)f(x) 有原函数 ⇒\Rightarrow 存在无限个原函数
  2. 任何两个原函数之差为常数
  3. 若 f(x)f(x) 连续 ⇍⇒\nLeftarrow\Rightarrow 存在原函数
  4. 不定积分 - 设 F(x)F(x) 为 f(x)f(x) 的一个原函数，F(x)+CF(x)+C 为所有原函数，成为不定积分，记为 ∫f(x)⋅dx=F(x)+C\displaystyle{\int f(x)\cdot dx = F(x)+C}

不定积分工具

基本公式

  1. ∫k⋅dx=kx+C\displaystyle{\int k\cdot dx = kx + C}
  2. {∫xa⋅dx=1a+1xa+1+Ca≠−1∫1x⋅dx=ln⁡∣x∣+C\begin{cases}\displaystyle{\int x^{a}\cdot dx = \frac{1}{a+1}x^{a+1} + C \quad a\neq -1}\\\displaystyle{\int \frac{1}{x}\cdot dx = \ln\vert x\vert + C}\end{cases}
  3. ∫ax⋅dx=axln⁡a+C\displaystyle{\int a^{x}\cdot dx = \frac{a^{x}}{\ln a} + C}
  4. 三角函数
    1. ∫sin⁡x⋅dx=−cos⁡x+C\displaystyle{\int \sin x\cdot dx = -\cos x+C}
    2. ∫cos⁡x⋅dx=sin⁡x+C\displaystyle{\int \cos x\cdot dx = \sin x+C}
    3. ∫tan⁡x⋅dx=−ln⁡∣cos⁡x∣+C\displaystyle{\int \tan x\cdot dx = -\ln \vert \cos x\vert+C}
    4. ∫cot⁡x⋅dx=ln⁡∣sin⁡x∣+C\displaystyle{\int \cot x\cdot dx = \ln \vert \sin x\vert+C}
    5. ∫sec⁡x⋅dx=ln⁡∣sec⁡x+tan⁡x∣+C\displaystyle{\int \sec x\cdot dx = \ln \vert \sec x + \tan x\vert+C}
    6. ∫csc⁡x⋅dx=ln⁡∣csc⁡x−cot⁡x∣+C\displaystyle{\int \csc x\cdot dx = \ln \vert \csc x - \cot x\vert+C}
    7. ∫sec⁡2x⋅dx=tan⁡x+C\displaystyle{\int \sec^{2} x\cdot dx = \tan x+C}
    8. ∫csc⁡2x⋅dx=−cot⁡x+C\displaystyle{\int \csc^{2} x\cdot dx = -\cot x+C}
    9. ∫sec⁡xtan⁡x⋅dx=sec⁡x+C\displaystyle{\int \sec x\tan x\cdot dx = \sec x + C}
    10. ∫csc⁡xcot⁡x⋅dx=−csc⁡x+C\displaystyle{\int \csc x\cot x\cdot dx = -\csc x + C}
  5. 反三角函数
    1. ∫11−x2⋅dx=arcsin⁡x+C\displaystyle{\int \frac{1}{\sqrt{1-x^{2}}} \cdot dx = \arcsin x + C}
    2. ∫1a2−x2⋅dx=arcsin⁡xa+C\displaystyle{\int \frac{1}{\sqrt{a^{2} - x^{2}}} \cdot dx = \arcsin \frac{x}{a} + C}
    3. ∫11+x2⋅dx=arctan⁡x+C\displaystyle{\int \frac{1}{1+x^{2}} \cdot dx = \arctan x + C}
    4. ∫1a2+x2⋅dx=1aarctan⁡xa+C\displaystyle{\int \frac{1}{a^{2} + x^{2}} \cdot dx = \frac{1}{a}\arctan \frac{x}{a} + C}
    5. ∫1x2+a2⋅dx=ln⁡(x+x2+a2)+C\displaystyle{\int \frac{1}{\sqrt{x^{2} + a^{2}}} \cdot dx = \ln(x+\sqrt{x^{2}+a^{2}}) + C}
    6. ∫1x2−a2⋅dx=ln⁡∣x+x2−a2∣+C\displaystyle{\int \frac{1}{\sqrt{x^{2}-a^{2}}} \cdot dx = \ln\vert x+ \sqrt{x^{2}-a^{2}} \vert + C}
    7. ∫1x2−a2⋅dx=12aln⁡∣x−ax+a∣+C\displaystyle{\int \frac{1}{x^{2}-a^{2}} \cdot dx = \frac{1}{2a}\ln\left\vert \frac{x-a}{x+a}\right\vert + C}
    8. ∫a2−x2⋅dx=a22arcsin⁡xa+x2a2−x2+C\displaystyle{\int \sqrt{a^{2}-x^{2}} \cdot dx = \frac{a^{2}}{2}\arcsin \frac{x}{a} + \frac{x}{2}\sqrt{a^{2}-x^{2}} + C}

积分法

换元积分法

第一类换元积分法

原理 ∫f[ϕ(x)]ϕ′(x)⋅dx=∫f[ϕ(x)]⋅dϕ(x)=ϕ(x)=t∫f(t)⋅dt=F(t)+C=F[ϕ(x)]+C\begin{split} &\int f[\phi(x)]\phi'(x)\cdot dx = \int f[\phi(x)]\cdot d\phi(x)\\ &\xlongequal{\phi(x) = t}\int f(t)\cdot dt = F(t) + C = F[\phi(x)] + C \end{split}

  • 例1 ∫xex2⋅dx\displaystyle{\int xe^{x^{2}}\cdot dx} ∫xex2⋅dx=∫ex212dx2=12∫ex2⋅dx2=12ex2+C \begin{split} \int_{}^{} xe^{x^{2}}\cdot dx = \int_{}^{}e^{x^{2}}\frac{1}{2}dx^{2} = \frac{1}{2}\int e^{x^{2}}\cdot dx^{2} = \frac{1}{2}e^{x^{2}} + C \end{split}
  • 例2 ∫x4+x4⋅dx\displaystyle{\int \frac{x}{4+x^{4}}\cdot dx} ∫x4+x4⋅dx=∫12⋅dx222+(x2)2=12⋅12arctan⁡x22+C=14arctan⁡x22+C \begin{split} &\int_{}^{}\frac{x}{4+x^{4}}\cdot dx = \int_{}^{}\frac{\frac{1}{2}\cdot dx^{2}}{2^{2}+(x^{2})^{2}} \\ = &\frac{1}{2}\cdot\frac{1}{2}\arctan\frac{x^{2}}{2} +C= \frac{1}{4}\arctan \frac{x^{2}}{2} + C \end{split}
  • 例3 ∫dxx⋅(1+x)\displaystyle{\int \frac{dx}{\sqrt{x}\cdot(1+x)}} ∫dxx(1+x)=∫2dx1+(x)2=2arctan⁡x+C \int_{}^{}\frac{dx}{\sqrt{x}(1+x)} = \int_{}^{}\frac{2 d \sqrt{x}}{1+(\sqrt{x})^{2}} = 2\arctan \sqrt{x} + C
  • 例4 ∫x(2x+3)2⋅dx\displaystyle{\int \frac{x}{(2x+3)^{2}}}\cdot dx ∫x(2x+3)2⋅dx=∫(2x+3)−32(2x+3)2⋅dx=14∫(2x+3)−3(2x+3)2⋅d(2x+3)=14ln⁡∣2x+3∣+34⋅12x+3+C \begin{split} & \int_{}^{} \frac{x}{(2x+3)^{2}}\cdot dx = \int_{}^{} \frac{(2x+3)-3}{2(2x+3)^{2}}\cdot dx \\ = & \frac{1}{4}\int_{}^{}\frac{(2x+3)-3}{(2x+3)^{2}}\cdot d(2x+3) \\ = & \frac{1}{4}\ln\vert 2x+3\vert + \frac{3}{4}\cdot\frac{1}{2x+3} + C \end{split}

  • 例5 ∫x−1x2+2x+2⋅dx\displaystyle{\int \frac{x-1}{x^{2}+2x+2}}\cdot dx ∫x−1x2+2x+2⋅dx=12∫2x+2x2+2x+2⋅dx−∫2x2+2x+2⋅dx=12∫d(x2+2x+2)x2+2x+2−∫1(x+1)2+1⋅d(1+x)=12ln⁡(x2+2x+2)−arctan⁡(1+x)+c \begin{split} &\int_{}^{} \frac{x-1}{x^{2}+2x+2}\cdot dx \\ = & \frac{1}{2}\int_{}^{}\frac{2x+2}{x^{2}+2x+2}\cdot dx - \int_{}^{}\frac{2}{x^{2}+2x+2}\cdot dx \\ = & \frac{1}{2}\int_{}^{}\frac{d(x^{2}+2x+2)}{x^{2}+2x+2} - \int_{}^{}\frac{1}{(x+1)^{2}+1}\cdot d(1+x) \\ = & \frac{1}{2}\ln (x^{2}+2x+2) - \arctan(1+x) +c \end{split}

    第二类换元积分法

    原理 ∫f(x)⋅dx=x=ϕ(t)∫f[ϕ(t)]ϕ′(t)⋅dt=∫g(t)⋅dt=G(t)+C=G[ϕ−1(x)]+C\begin{split}& \int f(x)\cdot dx \xlongequal{x = \phi(t)}\int f[\phi(t)]\phi'(t)\cdot dt = \int g(t)\cdot dt \\& = G(t)+C = G[\phi^{-1}(x)] + C \end{split}

    1. 无理化有理化
    2. 平方和平方差三角代换
      1. a2−x2=x=asin⁡ta⋅cos⁡t\displaystyle{\sqrt{a^{2}-x^{2}}\xlongequal{x=a\sin t}a\cdot \cos t}
      2. a2+x2=x=atan⁡ta⋅sec⁡t\displaystyle{\sqrt{a^{2}+x^{2}}\xlongequal{x=a\tan t}a\cdot \sec t}
      3. x2−a2=x=asec⁡ta⋅tan⁡t\displaystyle{\sqrt{x^{2}-a^{2}}\xlongequal{x=a\sec t}a\cdot \tan t}

  • 例1 ∫dx1+x\displaystyle{\int \frac{dx}{1+\sqrt{x}}} ∫dx1+x=x=t2∫2t⋅dt1+t=2t−2ln⁡∣1+t∣+C=2x−2ln⁡(1+x)+C \begin{split} &\int \frac{dx}{1+\sqrt{x}}\xlongequal{x=t^{2}}\int\frac{2t\cdot dt}{1+t} = 2t - 2\ln\vert 1+t\vert + C \\ & = 2\sqrt{x} - 2\ln(1+\sqrt{x}) + C \end{split}

  • 例2 ∫dxx2⋅1−x2\displaystyle{\int \frac{dx}{x^{2}\cdot \sqrt{1-x^{2}}}} ∫dxx2⋅1−x2=x=sin⁡t∫cos⁡tsin⁡2tcos⁡t⋅dt=∫1sin⁡2t⋅dt=−1a2cot⁡t+C=−1−x2x+C \begin{split} &\int \frac{dx}{x^{2}\cdot \sqrt{1-x^{2}}} \xlongequal{x = \sin t} \int \frac{\cos t}{\sin^{2} t\cos t}\cdot dt \\ & = \int \frac{1}{\sin^{2} t}\cdot dt = -\frac{1}{a^{2}}\cot t +C = -\frac{\sqrt{1-x^{2}}}{x} + C \end{split}
  • 例3 ∫dx(x2+1)3\displaystyle{\int \frac{dx}{\sqrt{(x^{2}+1)^{3}}}} ∫dx(x2+1)3=x=tan⁡t∫sec⁡2tsec⁡3t⋅dt=∫cos⁡t⋅dt=sin⁡t+C=x1+x2+C \begin{split} & \int \frac{dx}{\sqrt{(x^{2}+1)^{3}}} \xlongequal{x = \tan t} \int\frac{\sec^{2} t}{\sec^{3} t}\cdot dt = \int \cos t \cdot dt \\ & = \sin t + C = \frac{x}{\sqrt{1+x^{2}}} + C \end{split}

分部积分

基本思想 (uv)′=u′v+uv′⇒∫(uv)′⋅dx=∫u′v⋅dx+∫uv′⋅dx⇒uv=∫v⋅du+∫u⋅dv⇒∫u⋅dv=uv−∫v⋅du\begin{split}&(uv)' = u'v + uv'\\\Rightarrow & \int (uv)'\cdot dx = \int u'v\cdot dx + \int uv'\cdot dx \\\Rightarrow & uv = \int v \cdot du +\int u\cdot dv \\\Rightarrow & \int u\cdot dv = uv - \int v\cdot du\end{split}

∫幂×指数⋅dx\displaystyle{\int \text{幂}\times \text{指数}}\cdot dx

  • 例 ∫x2ex⋅dx\displaystyle{\int x^{2}e^{x}\cdot dx} ∫x2ex⋅dx=∫x2dex=x2ex−2∫xex⋅dx=x2ex−2∫x⋅dex=x2ex−2(xex−∫ex⋅dx)=x2ex−2xex+2ex+C \begin{split} & \int x^{2}e^{x}\cdot dx = \int x^{2}d e^{x} = x^{2}e^{x} -2\int xe^{x}\cdot dx \\ = & x^{2}e^{x} -2\int x\cdot de^{x} = x^{2}e^{x}-2(xe^{x} - \int e^{x}\cdot dx) \\ = & x^{2}e^{x} - 2xe^{x} + 2e^{x} + C \end{split}

∫幂×对数⋅dx\displaystyle{\int \text{幂}\times \text{对数}}\cdot dx

  • 例 ∫xln⁡2x⋅dx\displaystyle{\int x\ln^{2} x\cdot dx} ∫xln⁡2x⋅dx=12∫ln⁡2x⋅dx2=12(x2ln⁡2x−2∫xln⁡x⋅dx)=12x2ln⁡2x−12∫ln⁡x⋅dx2=12x2ln⁡2x−12(x2ln⁡x−∫x⋅dx)=12x2ln⁡2x−12x2ln⁡x+14x2+C \begin{split} & \int x\ln^{2} x\cdot dx = \frac{1}{2}\int_{}^{} \ln^{2} x \cdot dx^{2}\\ = & \frac{1}{2}(x^{2}\ln^{2} x - 2\int_{}^{}x \ln x \cdot dx) \\ = & \frac{1}{2}x^{2}\ln^{2}x - \frac{1}{2}\int_{}^{}\ln x\cdot dx^{2} \\ = & \frac{1}{2}x^{2}\ln^{2} x - \frac{1}{2}(x^{2}\ln x - \int_{}^{}x\cdot dx) \\ = & \frac{1}{2}x^{2}\ln^{2}x - \frac{1}{2}x^{2}\ln x + \frac{1}{4}x^{2} + C \end{split}

∫幂×三角⋅dx\displaystyle{\int \text{幂}\times \text{三角}}\cdot dx

  • 例1 ∫x2cos⁡2x⋅dx\displaystyle{\int x^{2}\cos 2x \cdot dx} ∫x2cos⁡2x⋅dx=12∫x2⋅dsin⁡2x=12(x2sin⁡2x−2∫xsin⁡2x⋅dx)=12x2sin⁡2x+12∫x⋅dcos⁡2x=12x2sin⁡2x+12(xcos⁡2x−∫cos⁡2x⋅dx)=12x2sin⁡2x+12xcos⁡2x−14sin⁡2x+C \begin{split} & \int_{}^{}x^{2}\cos 2x \cdot dx = \frac{1}{2}\int_{}^{}x^{2}\cdot d\sin 2x \\ = & \frac{1}{2}(x^{2}\sin 2x - 2\int_{}^{}x\sin 2x\cdot dx ) \\ = & \frac{1}{2}x^{2}\sin 2x + \frac{1}{2}\int_{}^{}x \cdot d\cos 2x \\ = & \frac{1}{2}x^{2}\sin 2x + \frac{1}{2}(x\cos 2x - \int_{}^{}\cos 2x\cdot dx) \\ = & \frac{1}{2}x^{2}\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x + C \end{split}
  • 例2 ∫xsin⁡2x⋅dx\displaystyle{\int x\sin^{2} x\cdot dx} ∫xsin⁡2x⋅dx=∫x(1−cos⁡2x2)⋅dx=∫x2⋅dx−12∫xcos⁡2x⋅dx=14x2−14∫x⋅dsin⁡2x=14x2−14(xsin⁡2x−∫sin⁡2x⋅dx)=14x2−14xsin⁡2x+18cos⁡2x+C \begin{split} & \int_{}^{}x\sin^{2} x\cdot dx = \int_{}^{}x\bigg(\frac{1-\cos 2x}{2}\bigg) \cdot dx \\ = & \int_{}^{}\frac{x}{2}\cdot dx - \frac{1}{2}\int_{}^{}x\cos 2x \cdot dx \\ = & \frac{1}{4}x^{2} - \frac{1}{4}\int_{}^{}x\cdot d\sin 2x \\ = & \frac{1}{4}x^{2} - \frac{1}{4}(x\sin 2x - \int \sin 2x \cdot dx) \\ = & \frac{1}{4}x^{2} - \frac{1}{4}x\sin 2x + \frac{1}{8}\cos 2x + C \end{split}
  • 例3 ∫x⋅tan⁡2x\displaystyle{\int x\cdot \tan^{2} x} ∫xtan⁡2x⋅dx=∫x(sec⁡2x−1)⋅dx=∫x⋅dtan⁡x−12x2=xtan⁡x−∫tan⁡x⋅dx−12x2=xtan⁡x+ln⁡∣cos⁡x∣−12x2+C \begin{split} & \int_{}^{} x\tan^{2} x \cdot dx = \int_{}^{}x(\sec^{2}x - 1)\cdot dx \\ = & \int_{}^{}x\cdot d \tan x - \frac{1}{2}x^{2} \\ = & x\tan x - \int \tan x \cdot dx - \frac{1}{2}x^{2} \\ = & x\tan x + \ln\vert \cos x\vert - \frac{1}{2}x^{2} + C \end{split}

∫幂×反三角⋅dx\displaystyle{\int \text{幂}\times \text{反三角}}\cdot dx

  • 例1 ∫x2arctan⁡x⋅dx\displaystyle{\int x^{2}\arctan x\cdot dx} ∫x2arctan⁡x⋅dx=13∫arctan⁡x⋅dx3=13(x3arctan⁡x−∫x31+x2⋅dx)=13x3arctan⁡x−13∫x(1+x2−x)1+x2⋅dx=13x3arctan⁡x−13∫x⋅dx+16∫dx21+x2=13x3arctan⁡x−16x2+16ln⁡(1+x2)+C \begin{split} & \int x^{2}\arctan x\cdot dx = \frac{1}{3}\int_{}^{} \arctan x \cdot d x^{3}\\ = & \frac{1}{3}(x^{3}\arctan x - \int_{}^{}\frac{x^{3}}{1+x^{2}}\cdot dx) \\ = & \frac{1}{3}x^{3}\arctan x - \frac{1}{3}\int_{}^{}\frac{x(1+x^{2}-x)}{1+x^{2}}\cdot dx \\ = & \frac{1}{3}x^{3}\arctan x - \frac{1}{3}\int x\cdot dx + \frac{1}{6}\int \frac{dx^{2}}{1+x^{2}} \\ = & \frac{1}{3}x^{3}\arctan x - \frac{1}{6}x^{2} + \frac{1}{6}\ln(1+x^{2}) +C \end{split}
  • 例2 ∫arcsin⁡x⋅dx\displaystyle{\int \arcsin x\cdot dx} ∫arcsin⁡x⋅dx=xarcsin⁡x−∫x1−x2⋅dx=xarcsin⁡x+12∫d(−x2)1−x2=xarcsin⁡x+1−x2+C \begin{split} & \int_{}^{}\arcsin x \cdot dx = x\arcsin x - \int_{}^{}\frac{x}{\sqrt{1-x^{2}}}\cdot dx \\ = & x\arcsin x + \frac{1}{2}\int_{}^{}\frac{d(-x^{2})}{\sqrt{1-x^{2}}} \\ = & x\arcsin x + \sqrt{1-x^{2}} + C \end{split}

∫eax×{cos⁡bxsin⁡bx⋅dx\displaystyle{\int e^{ax}\times \begin{cases}\cos bx\\\sin b x\end{cases}}\cdot dx

  • 例 ∫excos⁡2x⋅dx\displaystyle{\int e^{x}\cos 2x \cdot dx} 令I=∫excos⁡2x⋅dx∫excos⁡2x⋅dx=∫cos⁡2x⋅dex=excos⁡2x+2∫exsin⁡2x⋅dx=excos⁡2x+2(exsin⁡2x−2∫excos⁡2x⋅dx)=excos⁡2x+2exsin⁡2x−4I5I=excos⁡2x+2exsin⁡2xI=excos⁡2x+2exsin⁡2x5+C \begin{split} & \text{令}I = \int_{}^{}e^{x}\cos 2x\cdot dx \\ & \int e^{x}\cos 2x \cdot dx = \int_{}^{}\cos 2x \cdot de^{x} \\ = & e^{x}\cos 2x + 2\int_{}^{}e^{x}\sin 2x \cdot dx \\ = & e^{x}\cos 2x + 2(e^{x}\sin 2x - 2\int_{}^{}e^{x}\cos 2x \cdot dx) \\ = & e^{x}\cos 2x + 2e^{x}\sin 2x - 4I\\ &5I = e^{x}\cos 2x + 2e^{x}\sin 2x \\ & I = \frac{e^{x}\cos 2x+2e^{x}\sin 2x}{5} + C \end{split}

∫{sec⁡nxcsc⁡nx⋅dxn为奇数\displaystyle{\int \begin{cases}\sec^{n} x\\\csc^{n} x\end{cases}}\cdot dx \quad n\text{为奇数}

  • 例 ∫sec⁡3x⋅dx\displaystyle{\int \sec^{3} x\cdot dx} 令I=∫sec⁡3x⋅dxI=∫sec⁡x⋅dtan⁡x=sec⁡xtan⁡x−∫tan⁡2xsec⁡x⋅dx=sec⁡xtan⁡x−∫(sec⁡2x−1)sec⁡x⋅dx=sec⁡xtan⁡x−I+∫sec⁡x⋅dx=sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣+I2I=sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣I=12(sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣)+C \begin{split} & \text{令}I = \int_{}^{}\sec^{3}x\cdot dx \\ & I = \int_{}^{}\sec x \cdot d\tan x \\ = & \sec x \tan x - \int_{}^{}\tan^{2} x\sec x\cdot dx \\ = & \sec x \tan x - \int_{}^{}(\sec^{2}x -1)\sec x\cdot dx \\ = & \sec x\tan x-I+\int_{}^{}\sec x\cdot dx \\ = & \sec x\tan x + \ln\vert \sec x + \tan x\vert + I \\ & 2I = \sec x\tan x + \ln \vert \sec x + \tan x\vert \\ & I =\frac{1}{2}(\sec x\tan x+\ln\vert \sec x +\tan x\vert) + C \end{split}

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