不定积分的定义与性质

\[\int f(x)dx\]

  • 目标: 求出原函数.
  • 策略: 把复杂的积分化简为几个简单的积分.
  • 方法: 用性质, 换元法, 分部积分法, 其他技巧.
  • 重点: 能根据题目特点, 快速找到恰当的化简方法.

1 不定积分的定义

  • 是一族原函数. \[ \int f(x)dx=F(x)+C. \]

  • 满足关系\[F'(x)=f(x).\]

2 不定积分的性质

  • 积分与微分互逆. \[\begin{aligned} \left[\int f(x)dx\right]'=f(x),\quad d\int f(x)dx=f(x)dx.\\ \int f'(x)dx=f(x)+C,\quad \int df(x)=f(x)+C. \end{aligned}\]

\[\begin{aligned} \int kf(x)dx&=k\int f(x)dx,\\ \int [f(x)+g(x)]dx&=\int f(x)dx+\int g(x)dx. \end{aligned}\]

  • 可以把一个复杂积分拆成两个简单积分计算.

3 基本积分公式

  • \(C\) 为积分常数，\(a\)、\(k\) 为非零常数.

\[ \int k \, dx = kx + C. \] \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad (n \neq -1). \] \[ \int \frac{1}{x} \, dx = \ln|x| + C. \]

\[ \int e^x \, dx = e^x + C, \] \[ \int a^x \, dx = \frac{a^x}{\ln a} + C \quad (a > 0, a \neq 1). \]

  • 基本.

\[ \int \sin x \, dx = -\cos x + C, \] \[ \int \cos x \, dx = \sin x + C, \] \[ \int \tan x \, dx = \ln|\sec x| + C, \] \[ \int \cot x \, dx = -\ln|\csc x| + C, \]

\[ \int \sec x \, dx = \ln|\sec x + \tan x| + C, \] \[ \int \csc x \, dx = \ln|\csc x - \cot x| + C, \]

  • 补充

\[ \int \sec^2 x \, dx = \tan x + C, \] \[ \int \csc^2 x \, dx = -\cot x + C, \] \[ \int \sec x \tan x \, dx = \sec x + C, \] \[ \int \csc x \cot x \, dx = -\csc x + C. \]

\[ \int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin x + C = -\arccos x + C, \] \[ \int \frac{1}{1 + x^2} \, dx = \arctan x + C = -\text{arccot } x + C. \]

\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C, \] \[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a}\ln\left|\frac{x - a}{x + a}\right| + C, \]

\[ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C, \] \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C, \] \[ \int \frac{1}{\sqrt{x^2 \pm a^2}} \, dx = \ln\left|x + \sqrt{x^2 \pm a^2}\right| + C, \] \[ \int \sqrt{x^2 \pm a^2} \, dx = \frac{x}{2}\sqrt{x^2 \pm a^2} \pm \frac{a^2}{2}\ln\left|x + \sqrt{x^2 \pm a^2}\right| + C. \]