📝 题目
例 10 求 $\displaystyle{\int \frac{\mathrm{d}x}{{x}^{2} + {px} + q}}$ .
💡 答案与解析
解 分几种情形讨论.
情形 1 设二次三项式 ${x}^{2} + {px} + q$ 有两个不相等的实根 $\alpha$ 和 $\beta$ ,即
$$ {x}^{2} + {px} + q = \left( {x - \alpha }\right) \left( {x - \beta }\right) ,\;\alpha \neq \beta , $$
则有
$$ \int \frac{\mathrm{d}x}{{x}^{2} + {px} + q} = \int \frac{\mathrm{d}x}{\left( {x - \alpha }\right) \left( {x - \beta }\right) } $$
$$ = \frac{1}{\alpha - \beta }\left( {\int \frac{\mathrm{d}x}{x - \alpha }-\int \frac{\mathrm{d}x}{x - \beta }}\right) $$
$$ = \frac{1}{\alpha - \beta }\ln \left| \frac{x - \alpha }{x - \beta }\right| + C. $$
情形 2 设 ${x}^{2} + {px} + q$ 有重实根 $\gamma$ ,即
$$ {x}^{2} + {px} + q = {\left( x - \gamma \right) }^{2}. $$
这时有
$$ \int \frac{\mathrm{d}x}{{x}^{2} + {px} + q} = \int \frac{\mathrm{d}x}{{\left( x - \gamma \right) }^{2}} $$
$$ = - \frac{1}{x - \gamma } + C\text{ . } $$
情形 3 设 ${x}^{2} + {px} + q$ 有一对共轭复根 $\lambda \pm \mathrm{i}\mu$ ,这时
$$ {x}^{2} + {px} + q = {\left( x + \frac{p}{2}\right) }^{2} + q - \frac{{p}^{2}}{4} = {\left( x - \lambda \right) }^{2} + {\mu }^{2}, $$
其中 $\lambda = - \frac{p}{2},\mu = \sqrt{q - \frac{{p}^{2}}{4}}$ . 对这一情形有
$$ \int \frac{\mathrm{d}x}{{x}^{2} + {px} + q} = \int \frac{\mathrm{d}x}{{\left( x - \lambda \right) }^{2} + {\mu }^{2}} $$
$$ = \frac{1}{\mu }\arctan \frac{x - \lambda }{\mu } + C $$
$$ = \frac{1}{\sqrt{q - \frac{{p}^{2}}{4}}}\arctan \frac{x + \frac{p}{2}}{\sqrt{q - \frac{{p}^{2}}{4}}} + C. $$