Section02_特定函数的不定积分

$\displaystyle{\int R(x)\cdot dx}$

定义

• $R(x) = \frac{P(x)}{Q(x)}$，其中$P(x)$，$Q(x)$为多项式
  • 若 $\deg P < \deg Q$，$R(x)$为真分式
  • 若 $\deg P\ge \deg Q$，$R(x)$为假分式

原则

1. 原则一 若$R(x)$为假分式，$R(x) = \text{多项式}+\text{真分式}$，方法如下图所示：
2. 原则二 若 $R(x)$ 为真分式$\begin{cases}\text{分子不动}\\\text{分母因式分解}\end{cases}\Rightarrow \text{拆分为部分之和}$，例
   1. $\displaystyle{\frac{3x-2}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1}}$
   2. $\displaystyle{\frac{x^{3}-4x^{2}+1}{(x-2)(x+1)^{3}} = \frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}} + \frac{D}{(x+1)^{3}}}$
   3. $\displaystyle{\frac{x^{2}-3x+1}{(2x+1)(x^{2}+1)} = \frac{A}{2x +1} + \frac{Bx+C}{x^{2}+1}}$
   4. $\displaystyle{\frac{3x^{2}-1}{x^{2}(x^{2}+x+1)} = \frac{A}{x}+\frac{B}{x^{2}}+\frac{Cx+D}{x^{2}+x+1}}$

例题

• 例1 $\displaystyle{\int \frac{1}{x^{2}-x-12}\cdot dx}$ $$\begin{align} & \int_{}^{}\frac{1}{x^{2}-x-12}\cdot dx = \int_{}^{}\frac{1}{(x+3)(x-4)}\cdot dx \\ & = \frac{1}{7}\int_{}^{}\bigg(\frac{1}{x-4}-\frac{1}{x+3}\bigg)\cdot dx \\ & = \frac{1}{7}\ln\left\vert \frac{x-4}{x+3}\right\vert +C \end{align}$$
• 例2 $\displaystyle{\int \frac{dx}{x^{2}+2x+2}}$ $$\begin{align} & \int_{}^{}\frac{1}{x^{2}+2x+2}\cdot dx = \int_{}^{}\frac{1}{(x+1)^{2}+1} \cdot d(x+1) = \arctan (x+1) + C \end{align}$$
• 例3 $\displaystyle{\frac{5x-1}{x^{2}-x-2}\cdot dx}$ $$\begin{align} & \int_{}^{}\frac{5x-1}{x^{2}-x-2}\cdot dx = \int_{}^{}\frac{5x-1}{(x-2)(x+1)}\cdot dx \\ & = \int_{}^{}\bigg(\frac{3}{x-2}+\frac{2}{x+1}\bigg) \cdot dx \\ & = 3\ln\vert x-2\vert + 2\ln \vert x+1\vert +C \end{align}$$
• 例4 $\displaystyle{\int \frac{x+2}{x^{2}+x+1}}$ $$\begin{align} & \int_{}^{}\frac{x+2}{x^{2}+x+1}\cdot dx\\ & = \frac{1}{2}\int_{}^{}\frac{d(x^{2}+x+1)}{x^{2}+x+1} + \frac{3}{2}\int \frac{1}{(x+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}} \cdot d(x+\frac{1}{2})\\ & = \frac{1}{2}\ln(x^{2} + x + 1) + \frac{3}{2}\cdot\frac{2}{\sqrt{3}}\arctan \frac{2x+1}{\sqrt{3}} + C\\ \end{align}$$
• 例5 $\displaystyle{\int \frac{3x+2}{x(1+x^{2})}\cdot dx}$ $$\begin{align} & \int_{}^{}\frac{3x+2}{x(1+x^{2})}\cdot dx = \int_{}^{}\bigg(\frac{2}{x}+\frac{-2x+3}{1+x^{2}}\bigg)\cdot dx \\ & = 2\ln \vert x \vert + 3 \arctan x - \int_{}^{}\frac{2x}{1+x^{2}} \cdot dx \\ & = 2\ln \vert x \vert + 3 \arctan x - \ln(1+x^{2}) +C \end{align}$$
• 例6 $\displaystyle{\int \frac{x^{3}+3}{x^{2}(1+x)}\cdot dx}$ $$\begin{align} & \int_{}^{}\frac{x^{3}+3}{x^{2}(1+x)}\cdot dx = \int_{}^{} \frac{x^{2}(1+x) - x^{2}+3}{x^{2}(1+x)}\cdot dx \\ & = x + \int_{}^{}\bigg(\frac{-3x+3}{x^{2}}+\frac{2}{1+x}\bigg)\cdot dx \\ & = x -3\ln \vert x \vert -3 \frac{1}{x} + 2\ln \vert 1+x \vert +C \\ \end{align}$$
• 例7 $\displaystyle{\int \frac{dx}{x(x^{4}+3)}\cdot dx}$ $$\begin{align} & \int_{}^{} \frac{dx}{x(x^{4}+3)} \cdot dx = \int_{}^{}\frac{x^{3}}{x^{4}(x^{4}+3)}\cdot dx \\ & = \frac{1}{4}\int_{}^{}\frac{dx^{4}}{x^{4}(x^{4}+3)} = \frac{1}{12}\ln x^{4} - \frac{1}{12}\ln(x^{4}+3) \end{align}$$
• 例8 $\displaystyle{\int \frac{x^{2}+1}{x^{4}+1}\cdot dx}$ $$\begin{align} & \int_{}^{} \frac{x^{2}+1}{x^{4}+1} \cdot dx = \int_{}^{} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\cdot dx = \int_{}^{}\frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^{2}+(\sqrt{2})^{2}}\\ & = \frac{1}{\sqrt{2}}\arctan \frac{x-\frac{1}{x}}{\sqrt{2}} + C \end{align}$$

无理函数的不定积分

Case 1 无需化为有理

• 例1 $\displaystyle{\int \frac{dx}{\sqrt{x}(1+x)}}$ $$\text{原式} = 2\int \frac{d\sqrt{x}}{1+x} = 2\arctan \sqrt{x} + C$$
• 例2 $\displaystyle{\int \frac{dx}{\sqrt{x-x^{2}}}\cdot dx}$ $$\text{原式} = 2\int \frac{d\sqrt{x}}{\sqrt{1-x}} = 2\arcsin \sqrt{x} + C$$
• 例3 $\displaystyle{\int \sqrt{\frac{1+x}{1-x}}\cdot dx}$ $$\text{原式} = \int \frac{1+x}{\sqrt{1-x^{2}}}\cdot dx = \arcsin x - \frac{1}{2}\cdot 2\sqrt{1-x^{2}} + C$$

Case 2 化为有理

• 例1 $\displaystyle{\int e^{\sqrt{x}}\cdot dx}$ $$\begin{align} & \text{原式}\xlongequal{x=t^{2}} \int 2t\times e^{t}\cdot dt = \int 2t \cdot de^{t} \\ & = 2t e^{t} - \int 2e^{t}\cdot dt = 2te^{t}-2e^{t} +C\\ & = 2\sqrt{x}e^{\sqrt{x}} - 2e^{\sqrt{x}} +C \end{align}$$
• 例2 $\displaystyle{\int \frac{dx}{1+\sqrt{x}}}$ $$\begin{align} &\text{原式} \xlongequal{x=t^{2}} \int \frac{2t}{1+t}\cdot dt = \int \bigg(2 - \frac{2}{1+t}\bigg)\cdot dt \\ & = 2t -2\ln\vert 1+t\vert + C \\ & = 2\sqrt{x} -2\ln(1+\sqrt {x}) +C \end{align}$$

三角有理函数的不定积分

• 例1 $\displaystyle{\int \frac{dx}{1+\cos x}\cdot dx}$ $$\text{原式} =$$