待定法分解有理函数

#personal/blog #math/高等数学/不定积分

如果 $\frac{g (x)}{f (x)}$ 是有理真分式，那么有：

\begin{aligned} \frac{g (x)}{f (x)} & = \frac{g (x)}{a {(x - a_{1})}^{k_{1}} \cdot \cdot \cdot (x - a_{s})^{k_{s}} (x^{2} + p_{1} x + q_{1})^{l_{1}} \cdot \cdot \cdot (x^{2} + p_{t} x + q_{t})^{l_{t}}} \\ = [\frac{A_{11}}{x - a_{1}} + \frac{A_{21}}{{(x - a_{1})}^{2}} + \dots + \frac{A_{k_{1} 1}}{{(x - a_{1})}^{k_{i}}}] + \dots + [\frac{A_{1 s}}{x - a_{s}} + \frac{A_{2 s}}{{(x - a_{s})}^{2}} + \dots + \frac{A_{m s}}{{(x - a_{s})}^{k_{s}}}] \\ + [\frac{M_{11} x + N_{11}}{x^{2} + p_{1} x + q_{1}} + \frac{M_{21} x + N_{21}}{{(x^{2} + p_{1} x + q_{1})}^{2}} + \dots + \frac{M_{l_{1} 1} x + N_{l_{1} 1}}{{(x^{2} + p_{1} x + q_{1})}^{l_{1}}}] \\ + \dots + [\frac{M_{1 t} x + N_{1 t}}{x^{2} + p_{t} x + q_{t}} + \frac{M_{2 t} x + N_{2 t}}{{(x^{2} + p_{t} x + q_{t})}^{2}} + \dots + \frac{M_{l_{t}, t} x + N_{l_{t}, t}}{{(x^{2} + p_{t} x + q_{t})}^{l_{t}}}] \end{aligned}

例如对于该有理真分式，可以如此分解：

\frac{3 x + 6}{(x - 1)^{2} (x^{2} + x + 1)} = \frac{a}{x - 1} + \frac{b}{(x - 1)^{2}} + \frac{c x + d}{x^{2} + x + 1}

如此我们可以通过合并这些分式，通过解方程求出 $a, b, c$ ，如下：

\begin{aligned} 3 x + 6 & = a (x - 1) (x^{2} + x + 1) + b (x^{2} + x + 1) + (c x + d) (x - 1)^{2} \\ = (a + c) x^{3} + (b - 2 c + d) x^{2} + (b + c - 2 d) x + (b - a + d) \\ ⟹ {\begin{cases} a + c = 0 \\ b - 2 c + d = 0 \\ b + c - 2 d = 3 \\ b - a + d = 6 \end{cases} ⟹ {\begin{cases} a = - 2 \\ b = 3 \\ c = 2 \\ d = 1 \end{cases} \end{aligned}