有理函数不定积分

发表: 5/17/2026 更新: 5/26/2026 字数: 0 字时长: 0 分钟

$\begin{array}{r} R (x) = \frac{P (x)}{Q (x)} . w h e r e P (x) 、 Q (x) 为多项式 . \end{array}$

\deg P (x) \geq \deg Q (x) . R (x) - 假 分 式;

\deg P (x) < \deg Q (x) . R (x) - 真 分 式 .

对于 $\int R (x) d x :$
$1^{0} . I f R (x) 为假分式, 则 R (x) = 多项式 + 真分式 .$

$当 R (x) 为假分式, 则可使用多项式除法将 R (x) 分解为多项式 + 真分式 .$
$下面介绍多项式除法$
$\begin{array}{r} 例如 : \frac{x^{4} + 5 x^{3} - x + 4}{x^{2} - x - 2} \end{array}$
$\begin{array}{lr} x^{2} + 6 x + 8 \\ x^{2} - x - 2 & x^{4} + 5 x^{3} + 0 x^{2} - x + 4 \\ \underset{―}{x^{4} - x^{3} - 2 x^{2}} \\ 6 x^{3} + 2 x^{2} - x \\ \underset{―}{6 x^{3} - 6 x^{2} - 12 x} \\ 8 x^{2} + 11 x + 4 \\ \underset{―}{8 x^{2} - 8 x - 16} \\ 19 x + 20 \end{array}$
$\begin{array}{r} \Rightarrow \frac{x^{4} + 5 x^{3} - x + 4}{x^{2} - x - 2} = x^{2} + 6 x + 8 + \frac{19 x + 20}{x^{2} - x - 2} \end{array}$

$2^{0} . I f R (x) 为真分式 : R (x) = \frac{分子不变}{分母因式分解} = 部分和$
① $\begin{array}{r} R (x) = \frac{19 x + 20}{x^{2} - x - 2} = \frac{19 x + 20}{(x + 1) (x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2} \end{array}$
$由 A (x - 2) + B (x + 1) = 19 x + 20 \Rightarrow {\begin{aligned} A + B & = 19 \\ - 2 A + B & = 20 \end{aligned}$
② $\begin{array}{r} R (x) = \frac{x^{2} - 4 x + 11}{(x - 1)^{3} (2 x + 1)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{(x - 1)^{3}} + \frac{D}{2 x + 1} \end{array}$
③ $\begin{array}{r} R (x) = \frac{2 x^{3} + 1}{x^{2} (x^{2} - x + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C x + D}{x^{2} - x + 1} \end{array}$

例1.
(1)
$\begin{aligned} \int \frac{d x}{x^{2} - x - 6} & = \frac{d x}{(x - 3) (x + 2)} = \frac{1}{5} \int (\frac{1}{x - 3} - \frac{1}{x + 2}) d x \\ = \frac{1}{5} \ln | \frac{x - 3}{x + 2} | + C . \end{aligned}$
(2)
$\begin{aligned} \int \frac{d x}{x^{2} + x + 1} & = \int \frac{d (x + \frac{1}{2})}{(\frac{\sqrt{3}}{2})^{2} + (x + \frac{1}{2})^{2}} = \frac{1}{\frac{\sqrt{3}}{2}} \arctan (\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C \\ = \frac{2 \sqrt{3}}{3} \arctan (\frac{2 \sqrt{3}}{3} x + \frac{\sqrt{3}}{3}) + C . \end{aligned}$

例2.
(1) $\begin{array}{r} \int \frac{5 x + 1}{x^{2} - x - 2} d x \end{array}$
$\begin{array}{l} \begin{array}{r} 解 : \frac{5 x + 1}{x^{2} - x - 2} = \frac{5 x + 1}{(x + 1) (x - 2)} = \frac{A}{x - 2} + \frac{B}{x + 1} . \end{array} \\ 由 A (x + 1) + B (x - 2) = 5 x + 1 \\ \Rightarrow {\begin{aligned} A + B & = 5 \\ A - 2 B & = 1 \end{aligned} \Rightarrow {\begin{aligned} A & = \frac{11}{3} \\ B & = \frac{4}{3} \end{aligned} \\ \begin{array}{r} 原式 = \frac{11}{3} \ln | x - 2 | + \frac{4}{3} \ln | x + 1 | + C . \end{array} \end{array}$
(2)
$\begin{aligned} \int \frac{x - 1}{x^{2} + x + 1} d x & = \frac{1}{2} \int \frac{(2 x + 1) - 3}{x^{2} + x + 1} d x \\ = \frac{1}{2} \int \frac{d (x^{2} + x + 1)}{x^{2} + x + 1} - \frac{3}{2} \int \frac{d (x + \frac{1}{2})}{(\frac{\sqrt{3}}{2})^{2} + (x + \frac{1}{2})^{2}} \\ = \frac{1}{2} \ln (x^{2} + x + 1) - \sqrt{3} \arctan (\frac{2 \sqrt{3}}{3} x + \frac{\sqrt{3}}{3}) + C . \end{aligned}$

例3. $\begin{array}{r} \int \frac{2 x^{2} + 3}{(x - 1) (x^{2} + 1)} d x \end{array}$
$\begin{array}{l} \begin{array}{r} 解 : \frac{x^{2} + 3}{(x - 1) (x^{2} + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 1} . \end{array} \\ 由 A (x^{2} + 1) + (B x + C) (x - 1) = x^{2} + 3 \\ \Rightarrow {\begin{aligned} A + B & = 1 \\ C - B & = 0 \\ A - C & = 3 \end{aligned} \Rightarrow {\begin{aligned} A & = 2 \\ B & = - 1 \\ C & = - 1 \end{aligned} \\ \begin{aligned} 原式 & = 2 \ln | x - 1 | - \frac{1}{2} (\int \frac{2 x}{x^{2} + 1} d x + \int \frac{2}{x^{2} + 1} d x) \\ = 2 \ln | x - 1 | - \frac{1}{2} (\int \frac{d (x^{2} + 1)}{x^{2} + 1} + 2 \int \frac{d x}{x^{2} + 1}) \\ = 2 \ln | x - 1 | - \frac{1}{2} \ln (x^{2} + 1) - \arctan x + C . \end{aligned} \end{array}$

例4. $\begin{array}{r} \int \frac{d x}{x (x^{4} + 2)} \end{array}$

法一

$\begin{aligned} \int \frac{d x}{x (x^{4} + 2)} & = \int \frac{x^{3} d x}{x^{4} (x^{4} + 2)} = \frac{1}{4} \int \frac{d x^{4}}{x^{4} (x^{4} + 2)} \\ \overset{令 t = x^{4}}{=} \frac{1}{4} \int \frac{d t}{t (t + 2)} = \frac{1}{8} \int (\frac{1}{t} - \frac{1}{t + 2}) d t \\ = \frac{1}{8} \ln | \frac{t}{t + 2} | + C \\ = \frac{1}{8} \ln | \frac{x^{4}}{x^{4} + 2} | + C . \end{aligned}$

法二

$\begin{array}{l} \begin{array}{r} \int \frac{d x}{x (x^{4} + 2)} \overset{令 x = \sqrt{t}}{=} \int \frac{d (\sqrt{t})}{\sqrt{t} (t^{2} + 2)} = \frac{1}{2} \int \frac{d t}{t (t^{2} + 2)} \end{array} \\ \begin{array}{r} ∵ \frac{1}{t (t^{2} + 2)} = \frac{\frac{1}{2}}{t} + \frac{- \frac{1}{2} t}{t^{2} + 2} \end{array} \\ \begin{aligned} ∴ \frac{1}{2} \int \frac{d t}{t (t^{2} + 2)} & = \frac{1}{4} \int (\frac{1}{t} - \frac{t}{t^{2} + 2}) d t = \frac{1}{4} \int \frac{d t}{t} - \frac{1}{4} \int \frac{t}{t^{2} + 2} d t \\ = \frac{1}{4} \ln | t | - \frac{1}{8} \int \frac{d (t^{2} + 2)}{t^{2} + 2} = \frac{1}{8} \ln t^{2} - \frac{1}{8} \ln (t^{2} + 2) \\ = \frac{1}{8} \ln \frac{t^{2}}{t^{2} + 2} + C \\ = \frac{1}{8} \ln \frac{x^{4}}{x^{4} + 2} + C . \end{aligned} \end{array}$

例5.
(1)
$\begin{aligned} \int \frac{x^{2} + 1}{x^{4} + 1} d x & = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = \int \frac{d (x - \frac{1}{x})}{(\sqrt{2})^{2} + (x - \frac{1}{x})^{2}} \\ = \frac{1}{\sqrt{2}} \arctan (\frac{x - \frac{1}{x}}{\sqrt{2}}) + C . \end{aligned}$

(2)
$\begin{aligned} \int \frac{x^{2} - 1}{x^{4} + 1} d x & = \int \frac{1 - \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}}} d x = \int \frac{d (x + \frac{1}{x})}{(x + \frac{1}{x})^{2} - (\sqrt{2})^{2}} \\ = \frac{1}{2 \sqrt{2}} \ln | \frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}} | + C . \end{aligned}$

(3)
$\begin{array}{r} \int \frac{\sqrt{x}}{1 + x^{2}} d x \end{array}$

错题集

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有理函数不定积分