5.2 几类特殊函数的不定积分

1 有理函数

定义 有理函数 为 $f (x) = \frac{P_{n} (x)}{Q_{m} (x)},$ 它可以表示成多项式和真分式的和.

命题

有理函数的原函数是初等函数

证明

根据代数基本定理, 可以在 $R$ 内分解 $Q_{m} (x) = a_{0} (x - a)^{α} \dots (x - b)^{β} (x^{2} + p x + q)^{s} \dots (x^{2} + r x + h)^{t},$ 因此 $\frac{P_{n} (x)}{Q_{m} (x)}$ 可以拆分为 $\begin{aligned} \frac{P_{n} (x)}{Q_{m} (x)} = \frac{1}{a_{0}} & (\frac{A_{1}}{x - a} + \dots + \frac{A_{α}}{(x - a)^{α}} + \frac{B_{1}}{x - b} + \dots + \frac{B_{β}}{(x - b)^{β}} \\ + \dots + \frac{R_{1} x + Q_{1}}{x^{2} + p x + q} + \dots + \frac{R_{s} x + Q_{s}}{(x^{2} + p x + q)^{s}} \\ + \dots + \frac{C_{1} x + D_{1}}{x^{2} + r x + h} + \dots + \frac{C_{t} x + D_{t}}{(x^{2} + r x + h)^{t}} . \end{aligned}$
可以归类为四类不定积分:

$\int \frac{A}{x - a} d x = A \ln | x - a | + C$ .
$\int \frac{A}{(x - a)^{α}} d x = \frac{A}{1 - α} (x - a)^{1 - α} + C, α \neq 1$ .
回顾积分表第一条, $\begin{aligned} \int \frac{A x + B}{x^{2} + p x + q} d x = \frac{A}{2} \int \frac{2 x + p - p + \frac{2 B}{A}}{x^{2} + p x + q} d x \\ = & \frac{A}{2} \int \frac{d (x^{2} + p x + q)}{x^{2} + p x + q} - \frac{A}{2} (p - \frac{2 B}{A}) \int \frac{d x}{{(x + \frac{p}{2})}^{2} + (q - \frac{p^{2}}{4})} \\ = & \frac{A}{2} \ln | x^{2} + p x + q | - \frac{A}{2} (p - \frac{2 B}{A}) {(q - \frac{p^{2}}{4})}^{- \frac{1}{2}} \arctan \frac{x + \frac{p}{2}}{\sqrt{q - \frac{p^{2}}{4}}} + C . \end{aligned}$
$n \geq 2$ , $\begin{aligned} \int \frac{A x + B}{(x^{2} + p x + q)^{n}} d x \\ = & \frac{A}{2} \int \frac{d (x^{2} + p x + q)}{(x^{2} + p x + q)^{n}} - \frac{A}{2} (p - \frac{2 B}{A}) \int \frac{d x}{{[{(x + \frac{p}{2})}^{2} + (q - \frac{p^{2}}{4})]}^{n}} \\ = & \frac{A}{2} \int \frac{d (x^{2} + p x + q)}{(x^{2} + p x + q)^{n}} - \frac{A}{2} (p - \frac{2 B}{A}) \int \frac{d u}{(u^{2} + a^{2})^{n}} . \end{aligned}$ 记 $\begin{aligned} I_{n} & = \int \frac{d u}{(u^{2} + a^{2})^{n}} = \frac{u}{(u^{2} + a^{2})^{n}} - \int u d \frac{1}{(u^{2} + a^{2})^{n}} \\ = \frac{u}{(u^{2} + a^{2})^{n}} + \int \frac{u n (u^{2} + a^{2})^{n - 1} \cdot 2 u}{(u^{2} + a^{2})^{2 n}} d u \\ = \frac{u}{(u^{2} + a^{2})^{n}} + 2 n \int \frac{u^{2} + a^{2} - a^{2}}{(u^{2} + a^{2})^{n + 1}} d u \\ = \frac{u}{(u^{2} + a^{2})^{n}} + 2 n \int \frac{1}{(u^{2} + a^{2})^{n}} d u - 2 n a^{2} \int \frac{1}{(u^{2} + a^{2})^{n + 1}} d u \\ = \frac{u}{(u^{2} + a^{2})^{n}} + 2 n I_{n} - 2 n a^{2} I_{n + 1}, \end{aligned}$ (这里使用了分部积分法), 因此 $I_{n}$ 有通项 $I_{n + 1} = \frac{1}{2 n a^{2}} (\frac{u}{(u^{2} + a^{2})^{n}} + (2 n - 1) I_{n}) .$

综上, 结论得证.

当然, 这只是理论上的普适法则, 并不一定是最快做法.

例子

$\int \frac{x^{2} + 1}{x (x - 1)^{3}} d x$ . 尝试待定系数让 $x^{2} + 1 = A (x - 1)^{3} + B x (x - 1)^{2} + C x (x - 1) + D x,$ 得 $(A, B, C, D) = (- 1, 2, 1, 2)$ .
$\int \frac{x^{3}}{(x - 1)^{4}} d x = \int \frac{(t + 1)^{3}}{t^{4}} d t$ .
$\int \frac{d x}{x^{11} + 2 x} = \int \frac{x^{9}}{x^{20} + 2 x^{10}} d x = \frac{1}{10} \int \frac{d x^{10}}{x^{10} (x^{10} + 2)}$ .

2 三角有理函数

对 $\sin x, \cos x, \tan x, \cot x$ 施行四则运算得到的式子, 可表示为 $R (\sin x, \cos x)$ . 我们可以通过万能公式, 转换为有理函数的问题.

例子

$\int \frac{1 + \sin x}{\sin x (1 + \cos x)} d x = \int \frac{u^{2} + 2 u + 1}{2 u} d u$ , 这里直接应用万能公式, 令 $u = \tan \frac{x}{2}$ .
根据辅助角公式,

\begin{aligned} \int \frac{1}{a \sin x + b \cos x} d x = \int \frac{d x}{\sqrt{a^{2} + b^{2}} \sin (x + φ)} \\ = & \frac{1}{\sqrt{a^{2} + b^{2}}} \ln | \csc (x + φ) - \cot (x + φ) | + c . \end{aligned}

$\int \frac{c \sin x + d \cos x}{a \sin x + b \cos x} d x = \int \frac{k (a \sin x + b \cos x)^{'} + l (a \sin x + b \cos x)}{a \sin x + b \cos x} d x$ , 通过待定系数得到 $k, l$ .
$\int \frac{1}{a^{2} \sin^{2} x + b^{2} \cos^{2} x} d x = \int \frac{d \tan x}{a^{2} \tan^{2} x + b^{2}}$ .

3 简单无理函数

$\int R (x, (c x + b)^{\frac{n}{m}}) d x$ . 例如: $\begin{aligned} \int \frac{\sqrt{1 + x} + 2}{(x + 1)^{2} - \sqrt{x + 1}} d x = 2 \int \frac{t + 2}{t^{3} - 1} d t \\ = & 2 \int (\frac{1}{t - 1} - \frac{t + 1}{t^{2} + t + 1}) d t . \end{aligned}$ (令 $\sqrt{1 + x} = t$ )
$\int R (x, \sqrt[m]{\frac{a x + b}{c x + d}}) d x$ . 例如 $\begin{aligned} \int \frac{d x}{\sqrt[3]{(x - 1) (x + 1)^{2}}} = \int \frac{d x}{(x + 1) \sqrt[3]{\frac{x - 1}{x + 1}}} \\ = & \int \frac{\sqrt[3]{\frac{x + 1}{x - 1}}}{x + 1} d x = \int \frac{- 3}{t^{3} - 1} d t, \end{aligned}$ 这里 $t = \sqrt[3]{\frac{x + 1}{x - 1}}$ .
$\int R (x, \sqrt{a x^{2} + b x + c}) d x$ . 例如 $\int \frac{x + 1}{\sqrt{x^{2} + 6 x + 5}} d x = \int \frac{x + 1}{\sqrt{(x + 3)^{2} - 4}} d x = 2 \int (\sec^{2} u - \sec u) d u,$ 这里 $x + 3 = 2 \sec u$ .

例子

令 $x = \ln (1 + t^{2})$ , $\begin{aligned} \int \frac{x e^{x}}{\sqrt{e^{x} - 1}} d x = & 2 \int x d \sqrt{e^{x} - 1} = 2 x \sqrt{e^{x} - 1} - 2 \int \sqrt{e^{x} - 1} d x \\ = & 2 x \sqrt{e^{x} - 1} - 4 \int \frac{t^{2}}{1 + t^{2}} d t . \end{aligned}$
$\int \frac{\arctan x}{x^{2} (x^{2} + 1)} d x = \int (\frac{\arctan x}{x^{2}} - \frac{\arctan x}{x^{2} + 1}) d x$ .
$\int \frac{1 - \ln x}{(x - \ln x)^{2}} d x = \int {(1 - \frac{\ln x}{x})}^{- 2} d \frac{\ln x}{x}$ .

\begin{aligned} \int \frac{1 + \sin x}{1 + \cos x} e^{x} d x = \int \frac{e^{x}}{\cos^{2} \frac{x}{2}} d \frac{x}{2} + \int \frac{\sin \frac{x}{2}}{\cos^{2} \frac{x}{2}} e^{x} d x \\ = & \int e^{x} d \tan \frac{x}{2} + \int \tan \frac{x}{2} e^{x} d x = e^{x} \tan \frac{x}{2} + C . \end{aligned}