MIT_单变量微积分_26

1.部分分式

Ex1: $$\begin{gathered} \int (\frac{1}{x+1}+\frac{3}{x+2})dx \\ =ln|x-1|+3ln|x+2|+C \end{gathered}$$

$$\begin{gathered} \frac{1}{x-1}+\frac{3}{x+2}=\frac{x+2+3(x-1)}{(x-1)(x+2)}=\frac{4x-1}{(x-1)(x+2)}=\frac{4x-1}{x^2+x-2} \\ 反推,\frac{4x-1}{x^2+x-2}=\frac{4x-1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} \end{gathered}$$

$$\begin{gathered} 求A:式子两边同时乘以x-1,得\frac{4x-1}{x+2}=A+\frac{B}{x+2}(x-1),令x=1，则A=1； \\ 求B，式子两边同时乘x+2,得\frac{4x-1}{x-1}=\frac{A(x+2)}{x-1}+B,令x=-2,则B=3； \end{gathered}$$
注意：带入数字时，不能带入相同的。
Ex2:$$\frac{x^2+2}{(x-1{)}^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}(\frac{7}{16}=\frac{0}{2}+\frac{4}{2^2}+\frac{2}{2^3}+\frac{1}{2^4})$$
$$\begin{gathered} 求C:令x=-2,同时乘x+2,\frac{x^2+2}{(x-1{)}^2}=\frac{A}{x-1}(x+2)+\frac{B}{(x-1{)}^2}(x+2)+C,C=\frac{6}{9}=\frac{2}{3} \\ 求B：令x=1,同时乘(x-1{)}^2,\frac{x^2+2}{x+2}=\frac{A}{x-1}(x-1{)}^2+B+\frac{C}{x+2}(x-1{)}^2,B=1; \\ 求A时：x≠=1，因此只能带入别的数，带入x=0,则\frac{2}{2}=\frac{A}{-1}+1+\frac{\frac{2}{3}}{2} \end{gathered}$$

Ex3: 分子最高幂次方小于分母高。

$$\begin{gathered} \frac{x^2}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1} \\ 求A,同乘x-1,\frac{x^2}{x^2+1}=A+\frac{Bx+C}{x^2+1}(x-1) \\ 令x=1,则A=\frac{1}{2} \\ 等式左边去分母： \\ {x}^2=A(x^2+1)+(Bx+C)(x-1) \\ {x}^2的系数：1=A(\frac{1}{2})+B,得B=\frac{1}{2} \\ 常数项:0=A(\frac{1}{2})-C得C=\frac{1}{2} \end{gathered}$$

Ex4: 分子幂次方比分母高。
$$\begin{gathered} \int \frac{x^2}{(x-1)(x^2+1)}dx=\int \frac{\frac{1}{2}}{x-1}dx+\int \frac{\frac{1}{2}x}{x^2+1}dx+\int \frac{\frac{1}{2}}{x^2+1}dx \\ =\frac{1}{2}ln|x-1|+\frac{1}{4}ln|x^2+1|+C \end{gathered}$$
$$\frac{x^3}{(x-1)(x+2)}=\frac{x^3}{x^2+x-2}=x-1+\frac{3x-2}{x^2+x-2}$$
下图求商、求余数的方法。

1.部分积分总结：

(1) $$\begin{gathered} \frac{P(x)}{Q(x)}=quotient(商)+\frac{P(x)}{Q(x)}(余数) \\ \frac{R(x)(最高阶小于12)}{(x+2{)}^4(x^2+2x+3)(x^2+4{)}^3}=\frac{A_1}{x+2}+\frac{A_2}{(x+2{)}^2}+\frac{A_3}{(x+2{)}^3}+\frac{A_4}{(x+2{)}^4} \\ +\frac{B_{0}x+C_0}{x^2+2x+3}+\frac{B_{1}x+C_1}{x^2+4}+\frac{B_{2}x+C_2}{(x^2+4{)}^2}+\frac{B_{3}x+C_3}{(x^2+4{)}^2} \\ =12个未知数 \end{gathered}$$
(2)对上述式子进行积分时，分别处理各个部分：
e.g:$$\int \frac{x}{(x^2+4{)}^3}dx=-\frac{1}{4}(x^2+4{)}^{-2}+C$$

e.g:$$\begin{gathered} \int \frac{dx}{(x^2+4{)}^3} \\ 令x=2tanu,dx=2se{c}^{2}du \\ =\int \frac{2se{c}^{2}u}{(4se{c}^{2}u{)}^3}du=\int \frac{du}{32se{c}^{4}u} \\ =\frac{1}{32}\int co{s}^{4}udu \\ =\frac{1}{32}\int (\frac{1+cos2u}{2}{)}^{2}du=... \end{gathered}$$
e.g:$$\int \frac{dx}{x^2+2x+3}=\int \frac{dx}{(x+1{)}^2+2}=ta{n}^{-1}\frac{x+1}{\sqrt{2}}$$