方企勤 第三章 一元函数积分学 第2题
📝 题目
例 2 求下列不定积分:
(1) $\displaystyle \int \frac{\mathrm{d}x}{{x}^{4}\left( {1 + {x}^{2}}\right) }\mathrm{d}x$ ; (2) $\displaystyle{\int \frac{\mathrm{d}x}{1 - {x}^{4}}\mathrm{\;d}x}$ .
💡 答案解析
解 (1) 原式 $= \int \frac{1 - {x}^{4} + {x}^{4}}{{x}^{4}\left( {1 + {x}^{2}}\right) }\mathrm{d}x = \int \frac{1 - {x}^{2}}{{x}^{4}}\mathrm{d}x + \int \frac{1}{1 + {x}^{2}}\mathrm{\;d}x$
$$ = \arctan x + \frac{1}{3{x}^{3}}\left( {3{x}^{2} - 1}\right) + C. $$
或 原式 $= \int \frac{1 + {x}^{2} - {x}^{2}}{{x}^{4}\left( {1 + {x}^{2}}\right) }\mathrm{d}x = \int \frac{\mathrm{d}x}{{x}^{4}} - \int \frac{\mathrm{d}x}{{x}^{2}\left( {1 + {x}^{2}}\right) }$
$$ = - \frac{1}{3{x}^{3}} - \int \frac{1 + {x}^{2} - {x}^{2}}{{x}^{2}\left( {1 + {x}^{2}}\right) }\mathrm{d}x $$
$$ = - \frac{1}{3{x}^{3}} + \frac{1}{x} + \arctan x + C. $$
(2)原式 $\displaystyle{= \frac{1}{2}\int \left\lbrack {\frac{1}{1 - {x}^{2}} + \frac{1}{1 + {x}^{2}}}\right\rbrack \mathrm{d}x}$
$$ = \frac{1}{4}\ln \left| \frac{1 + x}{1 - x}\right| + \frac{1}{2}\arctan x + C. $$