有理函数的积分

笔记下载:有理函数的积分

1,有理函数:\(\displaystyle\quad f(x)=\frac{P(x)}{Q(x)}\),\(P(x)\),\(Q(x)\ \)都是多项式

2,有理函数的积分

(1)化成真分式

(2)真分式化成部分分式

3,部分分式,\(\displaystyle\quad\frac{P(x)}{Q(x)}\quad\)是真分式

(1)\(Q(x)\)有单项因子\(\displaystyle\quad (x-a)\)

\(\displaystyle\quad\Rightarrow\quad\frac{P(x)}{Q(x)}\quad\)有部分分式\(\displaystyle\quad \frac{A}{x-a}\)

(2)\(Q(x)\)有重因子\(\displaystyle\quad\ (x-a)^m\)

\(\displaystyle\quad\Rightarrow\quad\frac{P(x)}{Q(x)}\quad\)有部分分式\(\displaystyle\quad \frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+ \cdots +\frac{A_m}{(x-a)^m}\)

(3)\(Q(x)\)有因子\(\displaystyle\quad x^2+px+q\),\(\displaystyle\quad p^2-4q<0\)

\(\displaystyle\quad\Rightarrow\quad\frac{P(x)}{Q(x)}\quad\)有部分分式\(\displaystyle\quad \frac{Ax+B}{x^2+px+q}\)

(4)\(Q(x)\)有重因子\(\displaystyle\quad \left(x^2+px+q\right)^m\)

\(\displaystyle\quad\Rightarrow\quad\frac{P(x)}{Q(x)}\quad\)有部分分式\(\displaystyle\quad \frac{A_1 X+B_1}{x^2+px+q}+\frac{A_2 X+B_2}{\left(x^2+px+q\right)^2}+\cdots+\frac{A_m X+B_m}{\left(x^2+px+q\right)^m}\)

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

解答:\(\displaystyle\quad\int \frac{x^2+2x-1}{2x^3+3x^2-2x}dx\)

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

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

\(\displaystyle\qquad=\int \left(\frac{A}{x}+\frac{B}{2x-1}+\frac{C}{x+2}\right)dx\)

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

\(\displaystyle\Longrightarrow\quad A(2x-1)(x+2)+Bx(x+2)+Cx(2x-1)=x^2+2x-1\)

当\(\ \displaystyle x=-2\),

上式为\(\displaystyle\quad C(-2)(-4-1)=4-4-1\quad\Rightarrow\quad 10C=-1\quad\Rightarrow\quad C=-\frac{1}{10}\)

当\(\ \displaystyle x=\frac{1}{2}\),

上式为\(\displaystyle\quad \frac{1}{2}B\cdot\frac{5}{2}=\frac{1}{4}+1-1\quad\Rightarrow\quad\frac{5}{4}B=\frac{1}{4}\quad\Rightarrow\quad B=\frac{1}{5}\)

当\(\ \displaystyle x=0\),

上式为\(\displaystyle\quad A(-2)=-1\quad\Rightarrow\quad A=\frac{1}{2}\)

因此,\(\displaystyle\quad\int \left(\frac{A}{x}+\frac{B}{2x-1}+\frac{C}{x+2}\right)dx\)

\(\displaystyle\qquad=\int \left(\frac{1}{2x}+\frac{1}{5}\cdot\frac{1}{2x-1}-\frac{1}{10}\cdot\frac{1}{x+2}\right)dx\)

\(\displaystyle\qquad=\frac{1}{2}ln|x|+\frac{1}{10}ln|2x-1|-\frac{1}{10}\ln|x+2|+C\)

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

解答:(1)将原积分化成真分式

\(\displaystyle\qquad\int \frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}dx\)

\(\displaystyle\quad=\int \left(x+1+\frac{4x}{x^3-x^2-x+1}\right)dx\)

\(\displaystyle\quad=\frac{1}{2}x^2+x+\int\frac{4x}{x^3-x^2-x+1}dx\)

又因\(\displaystyle\quad\frac{4x}{x^3-x^2-x+1}=\frac{4x}{x^2(x-1)-(x-1)}\)

\(\displaystyle\quad=\frac{4x}{(x-1)(x^2-1)}=\frac{4x}{(x-1)^2(x+1)}\)

\(\displaystyle\quad=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{\left(x-1\right)^2}\)

\(\displaystyle\Longrightarrow\quad A(x-1)^2+B(x-1)(x+1)+C(x+1)=4x\)

当\(\ x=-1\quad\Rightarrow\quad\ 4A=-4\quad\Rightarrow\quad A=-1\)

当\(\ x=1\quad\Rightarrow\quad\ 2C=4\quad\Rightarrow\quad C=2\)

当\(\ x=0\quad\Rightarrow\quad\ A-B+C=0\quad\Rightarrow\quad B=A+C=1\)

因此,原积分\(\displaystyle\quad=\frac{x^2}{2}+x+\int \left(-\frac{1}{x+1}+\frac{1}{x-1}+\frac{2}{\left(x-1\right)^2}\right)dx\)

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

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

解答:\(\displaystyle\quad\int \frac{2x^2-x+4}{x^3+4x}dx=\int \frac{2x^2-x+4}{x(x^2+4)}dx=\int \left(\frac{A}{x}+\frac{Bx+C}{x^2+4}\right)dx\)

由\(\displaystyle\quad A(x^2+4)+x(Bx+C)=2x^2-x+4\)

当\(\ x=0\quad\Rightarrow\quad 4A=4\quad\Rightarrow\quad A=1\)

\(\displaystyle\quad (x^2+4+Bx^2+Cx)=2x^2-x+4\quad\Rightarrow\quad B+1=2\quad\Rightarrow\quad B=1\)

\(\displaystyle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \quad\ \quad\ \ \Rightarrow\quad C=-1\)

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

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

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

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

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

\(\displaystyle\qquad\quad\ =\ln x+\frac{1}{2}\ln (x^2+4)-\frac{1}{2}\arctan \frac{x}{2}+C\)