不定积分的定义与性质

我们通过原函数的定义来定义不定积分。所谓一个函数的原函数就是这个函数是它的原函数的导数。不定积分就是一个函数的所有的原函数的集合。

笔记下载:不定积分的概念

1,已知 \(f(x)=F'(x)\quad\Longrightarrow\quad F(x)=?\)

\(F(x)\):\(f(x)\)的原函数

2,\(F(x)\)是\(f(x)\)的原函数\(\quad\Longrightarrow\quad\)\(F(x)+c\)也是\(f(x)\)的原函数

定义:\(\int f(x)dx=F(x)+c\quad\)(不定积分)

\(f(x)\):被积函数,\(x\):积分变量,\(\int\):积分号

3,定理:\(f(x)\)在区间上连续\(\quad\Longrightarrow\quad\)\(f(x)\)在此区间上有原函数,即不定积分存在

4,基本积分公式

(1)\(\displaystyle\int0dx=C\quad\Leftrightarrow\quad C’=0\)

(2)\(\displaystyle\int x^ndx=\frac{x^{n+1}}{n+1}\quad\Leftrightarrow\quad (x^n)’=n\cdot x^{n-1}\)

(3)\(\displaystyle\int\sin xdx=-\cos x+c\)

(4)\(\displaystyle\int\cos xdx=\sin x+c\)

(5)\(\displaystyle\int\sec^2 xdx=\tan x+c\)

(6)\(\displaystyle\int\csc^2 xdx=-\cot x+c\)

(7)\(\displaystyle\int\sec x\cdot\tan x dx=\sec x+c\)

(8)\(\displaystyle\int\csc x\cdot\cot x dx=-\csc x+c\)

(9)\(\displaystyle\int\frac{1}{x} dx= \ln|x|=c\)

(10)\(\displaystyle\int e^x dx=e^x+c\)

(11)\(\displaystyle\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x+c=-\arccos x+c\)

(12)\(\displaystyle\int \frac{dx}{1+x^2}=\arctan x+c=-\text{arccot} x+c\)

(13)\(\displaystyle\int a^xdx=\frac{a^x}{\ln a}+c\)

5,不定积分的性质

(1)\(\displaystyle\left(\int f(x)dx\right)’=f(x)\)

(2)\(\displaystyle\left(\int F(x)’dx\right)’=F(x)+C\)

(3)\(\displaystyle\int cf(x)dx=c\int f(x)dx\)

(4)\(\displaystyle\int[f(x)\pm g(x)]=\int f(x)dx\pm\int g(x)dx\)

例1,求积分\(\displaystyle\int\left(e^x-3\cos x + \frac{1}{1+x^2}\right)dx\)

解答:=\(\displaystyle\int e^xdx-3\int \cos xdx+\int \frac{dx}{1+x^2}\)

\(\qquad\ \ \) =\(\displaystyle e^x-3\sin x+\arctan x+C\)

例2,求积分\(\displaystyle\int\sqrt{x}(1-x+x^2)dx\)

解答:=\(\displaystyle\int\left(\sqrt{x}-x^{\frac{3}{2}}+x^{\frac{5}{2}}\right)dx\)\(\qquad\ \ \)=\(\displaystyle\int x^{\frac{1}{2}}dx-\int x^{\frac{3}{2}}dx+\int x^{\frac{5}{2}}dx\)

\(\qquad\ \ \)=\(\displaystyle\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+C\)

\(\qquad\ \ \)=\(\displaystyle\frac{2}{3}x^{\frac{3}{2}}-\frac{2}{5}x^{\frac{5}{2}}+\frac{2}{7}x^{\frac{7}{2}}+C\)

例3,求\(\displaystyle\int \frac{2x^4+x^2+3}{1+x^2}dx\)

解答: =\(\displaystyle\int \frac {2x^4+2x^2-x^2+4}{x^2+1}dx\)

\(\qquad\ \ \)=\(\displaystyle\int \left(2x^2-1+\frac{4}{x^2+1}\right)dx\)

\(\qquad\ \ \)=\(\displaystyle\frac{2}{3}x^3-x+4\arctan x+C\)