📝 题目
例 2 试求
$$ \frac{3{x}^{4} + 2{x}^{3} + 3{x}^{2} - 1}{\left( {x - 2}\right) {\left( {x}^{2} + 1\right) }^{2}} $$
的部分分式分解.
💡 答案与解析
解 设
$$ \frac{3{x}^{4} + 2{x}^{3} + 3{x}^{2} - 1}{\left( {x - 2}\right) {\left( {x}^{2} + 1\right) }^{2}} = \frac{A}{x - 2} + \frac{{Mx} + N}{{x}^{2} + 1} + \frac{{M}^{\prime }x + {N}^{\prime }}{{\left( {x}^{2} + 1\right) }^{2}}. $$
消去分母得
$$ 3{x}^{4} + 2{x}^{3} + 3{x}^{2} - 1 $$
$$ = A\left( {{x}^{4} + 2{x}^{2} + 1}\right) + \left( {{Mx} + N}\right) \left( {{x}^{3} - 2{x}^{2} + x - 2}\right) $$
$$ + \left( {{M}^{\prime }x + {N}^{\prime }}\right) \left( {x - 2}\right) \text{ . } $$
比较上式两边同次项的系数得
$$ \begin{array}{llll} {x}^{4} + A & + M & = 3, & \\ {x}^{3} & - {2M} + N & = 2, & \\ {x}^{2} - {2A} & + M & = 3, & \\ x & - {2M} + N & = 3, & \\ 1 & A & - {2N}\; + N & = 0, \\ & & & \end{array} $$
解该方程组得
$$ A = 3,\;M = 0,\;N = 2,\;{M}^{\prime } = 1,\;{N}^{\prime } = 0. $$
于是, 我们得到
$$ \frac{3{x}^{4} + 2{x}^{3} + 3{x}^{2} - 1}{\left( {x - 2}\right) {\left( {x}^{2} + 1\right) }^{2}} = \frac{3}{x - 2} + \frac{2}{{x}^{2} + 1} + \frac{x}{{\left( {x}^{2} + 1\right) }^{2}}. $$
通过部分分式分解, 求有理分式的不定积分的问题归结为计算以下两种类型积分:
I. $\displaystyle \int \frac{\mathrm{d}x}{{\left( x - a\right) }^{n}}$ ;
II. $\displaystyle \int \frac{{Mx} + N}{{\left( {x}^{2} + px + q\right) }^{n}}\mathrm{\;d}x$ .
我们已经会计算 I 型积分:
$$ \int \frac{\mathrm{d}x}{{\left( x - a\right) }^{n}} = \left\{ \begin{array}{ll} \frac{-1}{\left( {n - 1}\right) {\left( x - a\right) }^{n - 1}} + C, & n \neq 1, \\ \ln \left| {x - a}\right| + C, & n = 1. \end{array}\right. $$
为了计算 II 型积分,我们将 ${x}^{2} + {px} + q$ 写成
$$ {x}^{2} + {px} + q = {\left( x + p/2\right) }^{2} + q - {p}^{2}/4 = {t}^{2} + {b}^{2}, $$
这里 $t = x + p/2,b = \sqrt{q - {p}^{2}/4}$ . 通过变数替换 $x = t - p/2$ ,求 II 型积分的问题又归结为计算以下两种类型的不定积分:
$$ {\mathrm{{II}}}^{\prime }.\int \frac{t}{{\left( {t}^{2} + {b}^{2}\right) }^{n}}\mathrm{\;d}t; $$
$$ {\mathbb{I}}^{\prime \prime } \cdot \int \frac{\mathrm{d}t}{{\left( {t}^{2} + {b}^{2}\right) }^{n}}. $$
其中 ${\mathrm{{II}}}^{\prime }$ 型积分很容易计算:
$$ \int \frac{t\mathrm{\;d}t}{{\left( {t}^{2} + {b}^{2}\right) }^{n}} = \left\{ \begin{array}{ll} \frac{-1}{2\left( {n - 1}\right) {\left( {t}^{2} + {b}^{2}\right) }^{n - 1}} + C, & n \neq 1, \\ \frac{1}{2}\ln \left( {{t}^{2} + {b}^{2}}\right) + C, & n = 1. \end{array}\right. $$
为了计算 ${\mathrm{{II}}}^{\prime \prime }$ 型积分 ${J}_{n}$ ,可以利用已知的递推公式
$$ {J}_{n} = \frac{1}{2\left( {n - 1}\right) {b}^{2}} \cdot \frac{t}{{\left( {t}^{2} + {b}^{2}\right) }^{n - 1}} + \frac{{2n} - 3}{2\left( {n - 1}\right) {b}^{2}}{J}_{n - 1} $$
和已知的结果
$$ {J}_{1} = \frac{1}{b}\arctan \frac{t}{b} + C. $$
这样, 我们完全解决了求有理分式函数的不定积分的问题.