不定积分 - 高等数学笔记

四. 不定积分

定义 $F'(x)=f(x),x \in I$ $\mathrm{d}F(x)=f(x)\mathrm{d}x$ $F(x)$ $f(x)$ $I$ 上的原函数。

定理 $F'(x)=f(x)$ $f(x)$ $F(x)+C$

定义 $I$ $f(x)$ $f(x)$ $\int f(x)\mathrm{d}x$

$\begin{matrix} (1) & \int k d x = k x + C k 是常数 \end{matrix}$
$\begin{matrix} (2) & \int x^{μ} d x = \frac{x^{μ + 1}}{μ + 1} + C (μ \neq - 1) \end{matrix}$
$\begin{matrix} (3) & \int \frac{d x}{x} = \ln | x | + C \end{matrix}$
$\begin{matrix} (4) & \int \frac{d x}{1 + x^{2}} = \arctan x + C \end{matrix}$
$\begin{matrix} (5) & \int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C \end{matrix}$
$\begin{matrix} (6) & \int \cos x d x = \sin x + C \end{matrix}$
$\begin{matrix} (7) & \int \sin x d x = - \cos x + C \end{matrix}$
$\begin{matrix} (8) & \int \frac{d x}{\cos^{2} x} = \int \sec^{2} x d x = \tan x + C \end{matrix}$
$\begin{matrix} (9) & \int \frac{d x}{\sin^{2} x} = \int \csc^{2} x d x = - \cot x + C \end{matrix}$
$\begin{matrix} (10) & \int \sec x \tan x d x = \sec x + C \end{matrix}$
$\begin{matrix} (11) & \int \csc x \cot x d x = - \csc x + C \end{matrix}$
$\begin{matrix} (12) & \int e^{x} d x = e^{x} + C \end{matrix}$
$\begin{matrix} (13) & \int a^{x} d x = \frac{a^{x}}{\ln a} + C \end{matrix}$

线性性质 $f(x),\ g(x)$ $k$ 为非零参数

\int [f (x) + g (x)] d x = \int f (x) d x + \int g (x) d x

\int \sum_{i = 1}^{n} [k_{i} f_{i} (x)] d x = \sum_{i = 1}^{n} k_{i} \int f_{i} (x) d x

积分形式的不变性 $u=u(x)$ $x$ 的任一可微函数，

\int f (x) d x = F (x) + c \Leftrightarrow \int f (u) d u = F (u) + c

求导与积分运算的关系：

先积分后微分，两者互相抵消
$d \int f (x) d x = f (x) d x o r [\int f (x) d x]^{'} = f (x)$
先微分后积分，相互抵消后相差一个常数
$\int d F (x) = F (x) + c o r \int F^{'} (x) d x = F (X) + c$

$f(x)$ $I$ $f(x)$ $I$ $f(x)$ $I$ 上可积。

定理 $f(u)$ $u=\varphi(x)$ 可导，则有换元公式：

\int f [φ (x)] φ^{'} (x) d x = {[\int f (u) d u]}_{u = φ (x)}

$\int g(x)\mathrm{d}x$ $g(x)=f[\varphi(x)]\varphi'(x)\mathrm{d}x$ 的形式，那么

\int g (x) d x = {[\int f (u) d u]}_{u = φ (x)}

$f(u)$ 的积分。

常见形式：

定理 $x=\psi(t)$ $\psi'(t)\neq 0$ $f[\psi(t)]\psi'(t)$ 具有原函数，则有换元公式：

\int f (x) d x = {[\int f [ψ (t)] ψ^{'} (t) d t]}_{t = ψ^{- 1} (x)}

$\psi^{-1}(x)$ $x=\psi(t)$ 的反函数

方法：

根式代换
$\displaystyle{\int} R(x, \sqrt[n]{ax+b})\mathrm{d}x$ $\sqrt[n]{ax+b}=t$ .
$\displaystyle{\int}R\left(x, \sqrt[n]{\frac{ax+b}{cx+d}}\right)\mathrm{d}x$ $\sqrt[n]{\frac{ax+b}{cx+d}}=t$ .
三角代换
$\sqrt{a^2-x^2} \quad \sqrt{a^2+x^2} \quad\sqrt{x^2-a^2}$ 等二次因子，可作三角代换
$\sqrt{a^2-x^2} \Rightarrow x=a\sin t \Rightarrow a\cos t$
$\sqrt{a^2+x^2} \Rightarrow x=a\tan t \Rightarrow a\sec t$
$\sqrt{x^2-a^2} \Rightarrow x=a\sec t \Rightarrow a\tan t$
倒代换
$x^k\quad (k \in \mathbb{N})$
$\displaystyle{\int}\frac{p_n(x)}{x^k\sqrt{ax^2-bx+c}}\mathrm{d}x\quad a\neq 0,k \in \mathrm{N},p_n(x)$ $n$ $n<k$ $x=\frac{1}{t}$ $x^k$ .

$u=u(x),v=v(x)$ 可导，则：

\begin{matrix} \int u v^{'} d x = u v - \int u^{'} v d x \\ \int u d v = u v - \int v d u \end{matrix}

$\int v\mathrm{d}u$ $\int u\mathrm{d}v$ 的积分容易求出。

$\frac{P(x)}{Q(x)}$ $P(x)$ $Q(x)$ 时，称为真分式，否则为假分式

$\displaystyle{\int}\frac{a\sin x+b\cos x}{a_1\sin x+b_1\cos x}\mathrm{d}x$ 可设
$a \sin x + b \sin x = A (a_{1} \sin x + b_{1} \cos x) + B (a_{1} \sin x + b_{1} \cos x)^{'}$
$A, B$ 然后裂项积分。
$\sqrt[n]{ax+b}$ $\sqrt[n]{\frac{ax+b}{cx+d}}$ $u$

\begin{matrix} (1) & \begin{aligned} \int \sin x d x = - \cos x + C; \int \cos x d x = \sin x + C; \\ \int \tan x d x = - \ln | \cos x | + C; \int \cot x d x = \ln | \sin x | + C; \\ \int \frac{d x}{\cos x} = \int \sec x d x = \ln | \sec x + \tan x | + C; \\ \int \frac{d x}{\sin x} = \int \csc x d x = \ln | \csc x - \cot x | + C; \\ \int \sec^{2} x d x = \tan x + C; \int \csc^{2} x d x = - \cot x + C; \\ \int \sec x \tan x d x = \sec x + C; \int \csc x \cot x d x = - \csc x + C . \end{aligned} \end{matrix}

\begin{matrix} (2) & {\begin{cases} \int \frac{1}{1 + x^{2}} d x = \arctan x + C, \\ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \arctan \frac{x}{a} + C & (a > 0) . \end{cases} \end{matrix}

\begin{matrix} (3) & {\begin{cases} \int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C, \\ \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \arcsin \frac{x}{a} + C & (a > 0) . \end{cases} \end{matrix}

\begin{matrix} (4) & {\begin{cases} \int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = \ln (x + \sqrt{x^{2} + a^{2}}) + C & (常见 a = 1), \\ \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x = \ln | x + \sqrt{x^{2} - a^{2}} | + C & (| x | > | a |) . \end{cases} \end{matrix}

\begin{matrix} (5) & \int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C (\int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} \ln | \frac{x + a}{x - a} | + C) . \end{matrix}

\begin{matrix} (6) & \int \sqrt{a^{2} - x^{2}} d x = \frac{a^{2}}{2} \arcsin \frac{x}{a} + \frac{x}{2} \sqrt{a^{2} - x^{2}} + C (a > | x | ⩾ 0) . \end{matrix}

\begin{matrix} (7) & \begin{aligned} \int \sin^{2} x d x = \frac{x}{2} - \frac{\sin 2 x}{4} + C (\sin^{2} x = \frac{1 - \cos 2 x}{2}) \\ \int \cos^{2} x d x = \frac{x}{2} + \frac{\sin 2 x}{4} + C (\cos^{2} x = \frac{1 + \cos 2 x}{2}) \\ \int \tan^{2} x d x = \tan x - x + C (\tan^{2} x = \sec^{2} x - 1) \\ \int \cot^{2} x d x = - \cot x - x + C (\cot^{2} x = \csc^{2} x - 1) \end{aligned} \end{matrix}