📝 题目
例 9 求不定积分 $\displaystyle{I = \int \frac{\mathrm{d}x}{1 + {x}^{2} + {x}^{4}},J = \int \frac{{x}^{2}}{1 + {x}^{2} + {x}^{4}}\mathrm{\;d}x}$ .
💡 答案与解析
解 $\displaystyle{I + J = \int \frac{1 + {x}^{2}}{1 + {x}^{2} + {x}^{4}}\mathrm{\;d}x = \int \frac{1 + \frac{1}{{x}^{2}}}{{x}^{2} + \frac{1}{{x}^{2}} + 1}\mathrm{\;d}x}$
$$ = \int \frac{\mathrm{d}\left( {x - \frac{1}{x}}\right) }{{\left( x - \frac{1}{x}\right) }^{2} + 3} = \frac{1}{\sqrt{3}}\arctan \frac{{x}^{2} - 1}{\sqrt{3}x} + C. $$
(1.2)
$$ - I + J = \int \frac{{x}^{2} - 1}{1 + {x}^{2} + {x}^{4}}\mathrm{\;d}x = \int \frac{1 - \frac{1}{{x}^{2}}}{{x}^{2} + \frac{1}{{x}^{2}} + 1}\mathrm{\;d}x $$
$$ = \int \frac{\mathrm{d}\left( {x + \frac{1}{x}}\right) }{{\left( x + \frac{1}{x}\right) }^{2} - 1} = \frac{1}{2}\ln \left| \frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}\right| + C $$
$$ = \frac{1}{2}\ln \left| \frac{{x}^{2} - x + 1}{{x}^{2} + x + 1}\right| + C. \tag{1.3} $$
联立 (1.2) 与 (1.3) 式解得
$$ I = \frac{1}{2\sqrt{3}}\arctan \frac{{x}^{2} - 1}{\sqrt{3}x} - \frac{1}{4}\ln \left| \frac{{x}^{2} - x + 1}{{x}^{2} + x + 1}\right| + C, $$
$$ J = \frac{1}{2\sqrt{3}}\arctan \frac{{x}^{2} - 1}{\sqrt{3}x} + \frac{1}{4}\ln \left| \frac{{x}^{2} - x + 1}{{x}^{2} + x + 1}\right| + C. $$