📝 题目
例 6 求 $\displaystyle{\int \frac{{x}^{2} + {3ax} - 1}{{x}^{4} + {x}^{2} - 2}\mathrm{\;d}x}$ .
💡 答案与解析
解 我们有
$$ \frac{{x}^{2} + {3ax} - 1}{{x}^{4} + {x}^{2} - 2} = \frac{{3ax} + \left( {{x}^{2} - 1}\right) }{\left( {{x}^{2} - 1}\right) \left( {{x}^{2} + 2}\right) } $$
$$ = \frac{{ax}\left\lbrack {\left( {{x}^{2} + 2}\right) - \left( {{x}^{2} - 1}\right) }\right\rbrack + \left( {{x}^{2} - 1}\right) }{\left( {{x}^{2} - 1}\right) \left( {{x}^{2} + 2}\right) } $$
$$ = a\left( {\frac{x}{{x}^{2} - 1} - \frac{x}{{x}^{2} + 2}}\right) + \frac{1}{{x}^{2} + 2}. $$
于是得到
$$ \int \frac{{x}^{2} + {3ax} - 1}{{x}^{4} + {x}^{2} - 2}\mathrm{\;d}x = \frac{a}{2}\ln \left| \frac{{x}^{2} - 1}{{x}^{2} + 2}\right| + \frac{1}{\sqrt{2}}\arctan \frac{x}{\sqrt{2}} + C. $$