『学习笔记』不定积分

原函数：$F(x)’=f(x)$，$F(x)$ 为 $f(x)$ 的一个原函数。

原函数存在：$f(x)$ 在 $I$ 连续，$\exists $ 可导函数 $F(x)$，$F’(x)=f(x)$

$\left(F(x)+C\right)’ = f(x)$
$F’(x) = f(x)$，$\varPhi’(x) = f(x)$，$\left(F(x)-\varPhi(x)\right)’ = F’(x)-\varPhi’(x) = f(x)-f(x) \equiv 0$，$F(x)-\varPhi(x) \equiv C$

不定积分定义: $f(x)$ 的原函数的全体：记作：$\displaystyle\int {f(x)} dx$。若：$F’(x) = f(x)$，$\boxed{\displaystyle\int f(x) dx = F(x) + C}$

公式（显而易见）：

$\displaystyle \frac{d}{dx}\left(\int f(x)dx\right) = f(x)$
$\displaystyle d\left[\int f(x)dx\right] = f(x)dx$
$\displaystyle\int F’(x)dx = F(x) + C$
$\displaystyle\int 1 dF(x) = F(x) + C$

积分表：

$\displaystyle\int kdx = kx + C\longleftrightarrow \left(kx + C\right)’ = k$
$\displaystyle\int x^{\mu}dx = \frac{x^{\mu + 1}}{\mu + 1} + C$（$\mu \neq -1\longleftrightarrow \left(x^{\mu + 1}\right)’ = (\mu + 1)x^{\mu}$，$\left(\frac{x^{\mu + 1}}{\mu + 1}\right)’ = x^{\mu}$
$\displaystyle\int \frac{1}{x}dx = \ln|x| + C \longleftrightarrow (\ln|x| + C)’ = \frac{1}{x}$
$\displaystyle\int \frac{1}{1 + x^2}dx = \arctan x + C \longleftrightarrow (\arctan x + C)’ = \frac{1}{1 + x^2}$
$\displaystyle\int \frac{1}{\sqrt{1 - x^2}}dx = \arcsin x + C \longleftrightarrow (\arcsin x + C)’ = \frac{1}{\sqrt{1 - x^2}}$
$\displaystyle\int \cos x dx = \sin x + C \longleftrightarrow (\sin x + C)’ = \cos x$
$\displaystyle\int \sin x dx = -\cos x + C \longleftrightarrow (-\cos x + C)’ = \sin x$
$\displaystyle\int \frac{1}{\cos^2 x}dx = \int \sec^2 x dx = \tan x + C \longleftrightarrow (\tan x + C)’ = \sec^2 x = \frac{1}{\cos^2 x}$
$\displaystyle\int \frac{1}{\sin^2 x}dx = \int \csc^2 x dx = -\cot x + C \longleftrightarrow (-\cot x + C)’ = \csc^2 x = \frac{1}{\sin^2 x}$
$\displaystyle\int \sec x \tan x dx = \sec x + C \longleftrightarrow (\sec x + C)’ = \sec x \tan x$
$\displaystyle\int \csc x \cot x dx = -\csc x + C \longleftrightarrow (-\csc x + C)’ = \csc x \cot x$
$\displaystyle\int a^x dx = \frac{a^x}{\ln a} + C \longleftrightarrow \left(\frac{a^x}{\ln a} + C\right)’ = \frac{a^x \ln a}{\ln a} = a^x$
$\displaystyle\int e^x dx = e^x + C \longleftrightarrow (e^x + C)’ = e^x$

性质：

$\displaystyle\int \left[f(x) \pm g(x)\right] dx = \int f(x) dx \pm \int g(x) dx$
$\displaystyle \int k f(x) dx = k \int f(x) dx$

第一换元积分法：

$\displaystyle\int f(u) du \stackrel{u = \varphi(x)}{=} \int f(\varphi(x)) d\varphi(x) = \int f(\varphi(x)) \varphi’(x) dx$（求导）
$\displaystyle \int f(\varphi(x)) \varphi’(x) dx = \int f(\varphi(x)) d\varphi(x) = \int f(u) du$（还原）

通过第一换元积分法，得出：

$\displaystyle \int \tan x dx = \int \frac{\sin x}{\cos x} dx = -\int \frac{d\cos x}{\cos x} = -\ln|\cos x| + C$
$\displaystyle \int \cot x dx = \int \frac{\cos x}{\sin x} dx = \int \frac{d\sin x}{\sin x} = \ln|\sin x| + C$

例题 1：$\boxed{\displaystyle \int \frac{dx}{x(1 + 2\ln x)} }$

$\begin{align*} \int \frac{dx}{x(1 + 2\ln x)} &= \int \frac{1}{1 + 2\ln x} \cdot \frac{1}{x} dx = \int \frac{1}{1 + 2\ln x} d(\ln|x|) \quad (x > 0) \\ &= \frac{1}{2} \int \frac{1}{1 + 2\ln x} d(2\ln x + 1) = \frac{1}{2} \ln|2\ln x + 1| + C \end{align*}$

例题 2：$\boxed{\displaystyle \int \tan^5 x \sec^3 x \, dx }$

$\begin{align*} \int \tan^5 x \sec^3 x \, dx &= \int \tan^4 x \sec^2 x \cdot \sec x \tan x \, dx \int (\sec^2 x - 1)^2 \sec^2 x \, d(\sec x) \\ &= \int (\sec^6 x - 2\sec^4 x + \sec^2 x) \, d(\sec x) \\ &= \frac{\sec^7 x}{7} - \frac{2}{5} \sec^5 x + \frac{1}{3} \sec^3 x + C \end{align*}$

第二换元积分法：

$\displaystyle \int f(x)dx = \int f(\psi(t))\psi’(t)dt$，其中 $x = \psi(t)$，且 $\psi(t)$ 单调、可导。

例题：$\boxed{\displaystyle \int \sqrt{a^2 - x^2} dx(a > 0)}$：
令 $x = a\sin t$，$-\frac{\pi}{2} < t < \frac{\pi}{2}$，则 $dx = a\cos t dt$。代入原积分：

$\begin{align*} \int \sqrt{a^2 - x^2} dx&=\int \sqrt{a^2 - a^2\sin^2 t} \cdot a\cos t dt\\ &=a^2 \int \cos^2 t dt\\ &=a^2 \int \frac{\cos 2t + 1}{2} dt\\ &=\frac{a^2}{4}\sin 2t + \frac{a^2}{2}t + C \end{align*}$

回代，由 $\sin t = \frac{x}{a}$，得 $\sin 2t = 2 \cdot \frac{x}{a} \cdot \frac{\sqrt{a^2 - x^2}}{a}$，$t = \arcsin \frac{x}{a}$，则：

$\begin{align*} \frac{a^2}{4}\sin 2t + \frac{a^2}{2}t + C&=\frac{a^2}{4} \cdot 2 \cdot \frac{x}{a} \cdot \frac{\sqrt{a^2 - x^2}}{a} + \frac{a^2}{2} \arcsin \frac{x}{a} + C\\ &=\frac{1}{2}x\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin \frac{x}{a} + C \end{align*}$

常用三角换元技巧：

$\sqrt{a^2 - x^2}$，$x = a\sin t$
$\sqrt{x^2 + a^2}$，$x = a\tan t$
$\sqrt{x^2 - a^2}$，$x = \pm a\sec t$

简单的小公式：

$\displaystyle\int \sec x \, dx = \ln|\sec x + \tan x| + C$
$\displaystyle\int \csc x \, dx = \ln|\csc x - \cot x| + C$
$\displaystyle\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \arctan\frac{x}{a} + C$
$\displaystyle\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin\frac{x}{a} + C$
$\displaystyle\int \frac{1}{\sqrt{a^2 + x^2}} dx = \ln(x + \sqrt{x^2 + a^2}) + C$
$\displaystyle\int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C$

例题：$\boxed{\displaystyle\int \frac{dx}{\sqrt{1 + x - x^2}}}$

$\begin{align*} \displaystyle\int \frac{dx}{\sqrt{1 + x - x^2}} &= \int \frac{dx}{\sqrt{\frac{5}{4} - \left(x - \frac{1}{2}\right)^2}} = \int \frac{dx}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2}}\\ &= \int \frac{d\left(x - \frac{1}{2}\right)}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - \left(x - \frac{1}{2}\right)^2}} = \arcsin\frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}} + C = \arcsin\frac{2x - 1}{\sqrt{5}} + C\end{align*}$

分部积分法：

公式：$\displaystyle \int u dv=uv-\int vdu$

优先级：$e^x>\sin x>\cos x > x> x^2>x^n$

例题 1：$\boxed{\displaystyle\int e^x \sin 2x \, dx }$

$\begin{align*} \int e^x \sin 2x \, dx &= \int \sin 2x \, de^x = e^x \sin 2x - \int e^x \, d\sin 2x = e^x \sin 2x - 2\int e^x \cos 2x \, dx \\&= e^x \sin 2x - 2\int \cos 2x \, de^x = e^x \sin 2x - 2\left(e^x \cos 2x + 2\int e^x \sin 2x \, dx\right) \\&= e^x \sin 2x - 2e^x \cos 2x - 4\int e^x \sin 2x \, dx \\ \int e^x \sin 2x \, dx &= \frac{1}{5}e^x (\sin 2x - 2\cos 2x) + C \end{align*}$

例题 2（根号换元）：$\boxed{\displaystyle \int e^{\sqrt{x}} dx}$

$\begin{align*}&\text{令 } t = \sqrt{x}, x = t^2, dx = 2t dt \\ & \int e^{\sqrt{x}} dx=\int e^t \cdot 2t dt = 2\int t de^t = 2te^t - 2\int e^t dt = 2te^t - 2e^t + C \quad \text{（回代）} \\ &= 2\sqrt{x}e^{\sqrt{x}} - 2e^{\sqrt{x}} + C \end{align*}$

有理函数积分

有理函数：一个多项式分式，例 $\dfrac{x^2+2x+3}{x-5}$。

真分式：分母次数 $>$ 分子次数，例 $\dfrac{x-1}{x^2+2x-2}$
假分式：分母次数 $\le$ 分子次数，例 $\dfrac{x^3}{x^3+x^2-7}$

How: 假分式变成真分式？Answer：多项式除法（长除法）/ 拼凑法

例题 1：$\displaystyle\int \frac{3x + 2}{2x - 5} dx $

$\begin{align*} \int \frac{3x + 2}{2x - 5} dx &= \int \frac{3x - \frac{15}{2} + \frac{19}{2}}{2x - 5} dx = \int \left( \frac{3}{2} + \frac{19}{2} \cdot \frac{1}{2x - 5} \right) dx \\ &= \frac{3}{2}x + \frac{19}{2} \cdot \frac{1}{2} \int \frac{1}{2x - 5} d(2x - 5) \\ &= \frac{3}{2}x + \frac{19}{4} \ln|2x - 5| + C \end{align*}$

例题 2：$\displaystyle \int \frac{1}{(1 + 2x)(1 + x^2)}dx$

$\displaystyle\frac{A}{1 + 2x} + \frac{Bx + C}{x^2 + 1} = \frac{Ax^2 + A + Bx + C + 2Bx^2 + 2Cx}{(1 + 2x)(x^2 + 1)}$

$\begin{cases}
A + 2B = 0 \
B + 2C = 0 \
A + C = 1
\end{cases}$

$\displaystyle A = \frac{4}{5}$，$\displaystyle B= -\frac{2}{5}$，$\displaystyle C= \frac{1}{5}$

①：$\displaystyle\int\frac{\frac 4 5 }{1+2x}=\frac{2}{5}\int \frac{1}{1 + 2x}d(2x + 1) = \frac{2}{5}\ln|2x + 1| + c$

②：$\displaystyle\int \frac{-\frac{2}{5}x + \frac{1}{5}}{x^2 + 1}dx = -\frac{1}{5}\int \frac{2x - 1}{x^2 + 1}dx = -\frac{1}{5}\left(\int \frac{d(x^2 + 1)}{x^2 + 1} - \frac{1}{1 + x^2}\right)dx= -\frac{1}{5}\left(\ln(x^2 + 1) - \arctan x\right) + C$

$\displaystyle\int \frac{1}{(1 + 2x)(1 + x^2)}dx = \frac{2}{5}\ln|2x + 1| - \frac{1}{5}\ln(x^2 + 1) + \frac{1}{5}\arctan x + C$