微积分笔记（15）——不定积分与计算

Contents

求导的逆运算

原函数的概念

不定积分

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

$$\int f(x) \mathrm{d} x = F(x) + C$$

$$\int f^{\prime}(x) \mathrm{d} x = f(x) + C$$

$$\dfrac{\mathrm{d}}{\mathrm{d} x} \int f(x) \mathrm{d} x = f(x) \\ \mathrm{d} \int f(x) \mathrm{d} x = f(x) \mathrm{d} x$$

基本公式

$$\int \mathrm{d} x = \int 1 \mathrm{d} x = x +C,x \in \mathbb{R} \\ \int x^\alpha \mathrm{d} x = \dfrac{x^{\alpha + 1}}{\alpha + 1} + C,x \in (0,+\infty)(\alpha \not = -1) \\ \int \dfrac{1}{x} \mathrm{d} x = \int \dfrac{\mathrm {d} x}{x} = \ln |x| + C,x \not = 0(x \in (-\infty,0) \lor x \in (0,+\infty)) \\ \int e^x \mathrm{d} x = e^x + C,x \in \mathbb{R} \\ \int a^x \mathrm{d} x = \dfrac{a^x}{\ln a} + C,x \in \mathbb{R}(a > 0,a \not = 1) \\ \int \sin x \mathrm{d} x = -\cos x + C,\int \cos x \mathrm{d} x = \sin x + C,x \in \mathbb{R} \\ \int \dfrac{\mathrm{d} x}{\cos^2 x} = \tan x + C,|x – k \pi| < \dfrac{\pi}{2},k=0,\pm 1,\pm 2,\cdots \\ \int \dfrac{\mathrm{d} x}{1 + x^2} = \arctan x + C,x \in \mathbb{R} \\ \int \dfrac{\mathrm{d} x}{\sqrt{1 - x^2}} = \arcsin x +C,|x| < 1$$

线性性质

$$\int [\alpha f(x) + \beta g(x)] \mathrm{d} x = \alpha \int f(x) \mathrm{d} x + \beta \int g(x) \mathrm{d} x,\forall \alpha,\beta \in \mathbb{R}$$

分部积分与换元法——求不定积分的方法

回忆

$$u^{\prime}v = (uv)^{\prime} – uv^{\prime} \\ \therefore \int u^{\prime}v \mathrm{d} x = \int (uv)^{\prime} \mathrm{d} x – \int uv^{\prime} \mathrm{d} x = uv – \int uv^{\prime} \mathrm{d} x$$

分部积分公式

$$\int u^{\prime}(x) v(x) \mathrm{d} x = u(x)v(x) – \int u(x)v^{\prime}(x) \mathrm{d} x$$

$$\int v \mathrm{d} u = uv – \int u \mathrm{d} v$$

应用

1. 求 $\int \ln |x| \mathrm{d} x$，其中 $x > 0$ 或 $x < 0$。$$\int \ln |x| \mathrm{d} x = x \ln |x| - \int x \cdot \dfrac{1}{x} \mathrm{d} x = x \ln |x| - x + C \\$$

2. 求 $\int x e^x \mathrm{d} x$。

$$\int x e^x \mathrm{d} x = x e^x – \int 1 \cdot e^x \mathrm{d} x = (x – 1)e^x + C \\ \int x^2 e^x \mathrm{d} x = x^2 e^x – \int 2x \cdot e^x \mathrm{d} x = x^2 e^x – 2 \int x e^x \mathrm{d} x \\$$

3. 求 $\int e^x \cos x \mathrm{d} x$ 和 $\int e^x \sin x \mathrm{d} x$。

$$A = \int e^x \cos x \mathrm{d} x,B = \int e^x \sin x \mathrm{d} x \\ A = e^x \cos x + \int e^x \sin x \mathrm{d} x = e^x \cos x + B \\ B = e^x \sin x – \int e^x \cos x \mathrm{d} x = e^x \sin x – A \\ \therefore A = \dfrac{e^x}{2}(\cos x + \sin x) + C_1,B = \dfrac{e^x}{2}(\sin x – \sin x) + C_2$$

换元法（积分变量代换）

$$\int f(\varphi(x)) \varphi^{\prime}(x) \mathrm{d} x = \int f(u) \mathrm{d} u$$

1. 若 $F(u)$ 是 $f(u)$ 的一个原函数，则：
$$\int f(\varphi(x)) \varphi^{\prime}(x) \mathrm{d} x = F(\varphi(x)) + C$$

2. 若 $G(x)$ 是 $f(\varphi(x)) \varphi^{\prime}(x)$ 的一个原函数，则：
$$\int f(u) \mathrm{d} u = G(\varphi^{-1}(u)) + C$$

应用方法 1（凑微分）

$$\int f(\varphi(x)) \varphi^{\prime}(x) \mathrm{d} x = \int f(\varphi(x)) \mathrm{d} \varphi(x) \stackrel{u = \varphi(x)}{=} \int f(u) \mathrm{d} u = F(u) + C = F(\varphi(x)) + C$$

$$\int \dfrac{x}{1 + x^2} \mathrm{d} x = \int \dfrac{\frac{1}{2} \mathrm{d} (x^2 + 1)}{1 + x^2} \stackrel{u = 1 + x^2}{=} \int \dfrac{1}{2} \dfrac{\mathrm{d} u}{u} = \dfrac{1}{2} \ln |u| + C = \ln \sqrt{1 + x^2} + C \\ \int (x + a)^\alpha \mathrm{d} x = \dfrac{(x + a)^{\alpha + 1}}{1 + \alpha} + C \\ \int \dfrac{\mathrm{d} x}{a^2 + x^2} = \dfrac{1}{a} \arctan \dfrac{x}{a} + C \\ \int \dfrac{\mathrm{d} x}{\sqrt{a^2 – x^2}} = \arcsin \dfrac{x}{a} + C$$