📝 题目
例 7 求 $\displaystyle{\int \frac{\mathrm{d}x}{{x}^{4} + {x}^{2} + 1}}$ 和 $\displaystyle{\int \frac{\mathrm{d}x}{{x}^{4} + 1}}$ .
💡 答案与解析
解 我们有
$$ \int \frac{\mathrm{d}x}{{x}^{4} + {x}^{2} + 1} = \frac{1}{2}\int \frac{\left( {{x}^{2} + 1}\right) - \left( {{x}^{2} - 1}\right) }{{x}^{4} + {x}^{2} + 1}\mathrm{\;d}x $$
$$ = \frac{1}{2}\int \frac{{x}^{2} + 1}{{x}^{4} + {x}^{2} + 1}\mathrm{\;d}x - \frac{1}{2}\int \frac{{x}^{2} - 1}{{x}^{4} + {x}^{2} + 1}\mathrm{\;d}x $$
$$ = \frac{1}{2}\int \frac{1 + \frac{1}{{x}^{2}}}{{x}^{2} + 1 + \frac{1}{{x}^{2}}}\mathrm{\;d}x - \frac{1}{2}\int \frac{1 - \frac{1}{{x}^{2}}}{{x}^{2} + 1 + \frac{1}{{x}^{2}}}\mathrm{\;d}x $$
$$ = \frac{1}{2}\int \frac{\mathrm{d}\left( {x - \frac{1}{x}}\right) }{{\left( x - \frac{1}{x}\right) }^{2} + 3} - \frac{1}{2}\int \frac{\mathrm{d}\left( {x + \frac{1}{x}}\right) }{{\left( x + \frac{1}{x}\right) }^{2} - 1} $$
$$ = \frac{1}{2\sqrt{3}}\arctan \frac{{x}^{2} - 1}{\sqrt{3}x} + \frac{1}{4}\ln \frac{{x}^{2} + x + 1}{{x}^{2} - x + 1} + C. $$
用类似的办法可以求得
$$ \int \frac{\mathrm{d}x}{{x}^{4} + 1} = \frac{1}{2\sqrt{2}}\arctan \frac{{x}^{2} - 1}{\sqrt{2}x} + \frac{1}{4\sqrt{2}}\ln \frac{{x}^{2} + \sqrt{2}x + 1}{{x}^{2} - \sqrt{2}x + 1} + C. $$