📝 题目
例 5 求 $\displaystyle \int \frac{\mathrm{d}x}{{\left( {x}^{3} + 1\right) }^{2}}$ .
💡 答案与解析
解 因为 ${x}^{3} + 1 = \left( {x + 1}\right) \left( {{x}^{2} - x + 1}\right)$ ,所以可设
$$ \int \frac{\mathrm{d}x}{{\left( {x}^{3} + 1\right) }^{2}} = \frac{a{x}^{2} + {bx} + c}{{x}^{3} + 1} + \alpha \ln \left| {x + 1}\right| $$
$$ + \beta \ln \left( {{x}^{2} - x + 1}\right) $$
$$ + \frac{2\gamma }{\sqrt{3}}\arctan \frac{{2x} - 1}{\sqrt{3}} + C. $$
上式两边求导得
$$ \frac{1}{{\left( {x}^{3} + 1\right) }^{2}} = \frac{\left( {{2ax} + b}\right) \left( {{x}^{3} + 1}\right) - 3{x}^{2}\left( {a{x}^{2} + {bx} + c}\right) }{{\left( {x}^{3} + 1\right) }^{2}} $$
$$ + \frac{\alpha }{x + 1} + \beta \frac{{2x} - 1}{{x}^{2} - x + 1} + \frac{\gamma }{{x}^{2} - x + 1}. $$
以 ${\left( {x}^{3} + 1\right) }^{2}$ 乘上式两边得
$$ 1 = \left( {{2ax} + b}\right) \left( {{x}^{3} + 1}\right) - 3{x}^{2}\left( {a{x}^{2} + {bx} + c}\right) $$
$$ + \alpha \left( {{x}^{3} + 1}\right) \left( {{x}^{2} - x + 1}\right) + \beta \left( {{2x} - 1}\right) \left( {{x}^{3} + 1}\right) \left( {x + 1}\right) $$
$$ + \gamma \left( {{x}^{3} + 1}\right) \left( {x + 1}\right) \text{ . } $$
比较系数得:
$$ \begin{array}{l} {x}^{5} + 1 = 0, \\ {x}^{4} + 2 = 0, \\ {x}^{3} + 3 = 1, \\ {x}^{2} + 2 = 0, \\ 1 = 0, \\ \end{array} $$
解该方程组得
$$ a = 0,\;b = \frac{1}{3},\;c = 0, $$
$$ \alpha = \frac{2}{9},\;\beta = - \frac{1}{9},\;\gamma = \frac{1}{3}. $$
于是, 我们求得
$$ \int \frac{\mathrm{d}x}{{\left( {x}^{3} + 1\right) }^{2}} = \frac{x}{3\left( {{x}^{3} + 1}\right) } + \frac{1}{9}\ln \frac{{\left( x + 1\right) }^{2}}{{x}^{2} - x + 1} $$
$$ + \frac{2}{3\sqrt{3}}\arctan \frac{{2x} - 1}{\sqrt{3}} + C. $$
我们对有理分式的积分法做一小结. 任何有理分式都可写成整式与既约真分式之和. 为了积分既约真分式, 可以利用待定系数法将其写成简单分式之和. 简单分式的积分是我们已经知道的. 对某些情形,还可以采取灵活变通的办法做简单分式分解,并结合其他手段计算积分.