📝 题目
3. 1.10 求下列不定积分:
(1) $\displaystyle \int \frac{{2x} + 3}{\left( {x - 2}\right) \left( {x + 5}\right) }\mathrm{d}x$ ; (2) $\displaystyle{\int \frac{\mathrm{d}x}{8 - {2x} - {x}^{2}}}$ ;
(3) $\displaystyle \int \frac{\mathrm{d}x}{{\left( x + 1\right) }^{2}\left( {x - 1}\right) }$ ; (4) $\displaystyle{\int \frac{{2x} - 3}{{x}^{2} + {2x} + 1}\mathrm{\;d}x}$ ;
(5) $\displaystyle \int \frac{\mathrm{d}x}{\left( {x + 1}\right) \left( {{x}^{2} + 1}\right) }$ ; (6) $\displaystyle{\int \frac{{x}^{4}}{{x}^{4} + 5{x}^{2} + 4}\mathrm{\;d}x}$ .
💡 答案与解析
3. 1.10 (1) 由 $\frac{{2x} + 3}{\left( {x - 2}\right) \left( {x + 5}\right) } = \frac{1}{x - 2} + \frac{1}{x + 5}$ 得出
原式 $= \ln \left( {{3x} + {x}^{2} - {10}}\right) + C$ ;
(2) 由 $\frac{1}{8 - {2x} - {x}^{2}} = \frac{1}{6\left( {x + 4}\right) } - \frac{1}{6\left( {x - 2}\right) }$ 得出
$$ \text{ 原式 } = \frac{1}{6}\ln \left( {x + 4}\right) - \frac{1}{6}\ln \left( {x - 2}\right) + C\text{ ; } $$
(3) 由 $\frac{1}{{\left( x + 1\right) }^{2}\left( {x - 1}\right) } = \frac{1}{4\left( {x - 1}\right) } - \frac{1}{4\left( {x + 1}\right) } - \frac{1}{2{\left( x + 1\right) }^{2}}$ 得出
原式 $= \frac{1}{4}\ln \left( {x - 1}\right) - \frac{1}{4}\ln \left( {x + 1}\right) + \frac{1}{{2x} + 2} + C$ ;
(4) 由 $\frac{{2x} - 3}{{x}^{2} + {2x} + 1} = \frac{2}{x + 1} - \frac{5}{{\left( x + 1\right) }^{2}}$ 得出
$$ \text{ 原式 } = 2\ln \left( {x + 1}\right) + \frac{5}{x + 1} + C\text{ ; } $$
(5) 由 $\frac{1}{\left( {x + 1}\right) \left( {{x}^{2} + 1}\right) } = \frac{1}{2\left( {x + 1}\right) } + \frac{1}{{x}^{2} + 1}\left( {\frac{1}{2} - \frac{1}{2}x}\right)$ 得出原式 $= \frac{1}{2}\arctan x + \frac{1}{2}\ln \left( {x + 1}\right) - \frac{1}{4}\ln \left( {{x}^{2} + 1}\right) + C$ ; (6) 由 $\frac{{x}^{4}}{{x}^{4} + 5{x}^{2} + 4} = \frac{1}{3\left( {{x}^{2} + 1}\right) } - \frac{16}{3\left( {{x}^{2} + 4}\right) } + 1$ 得出
原式 $= x + \frac{1}{3}\arctan x - \frac{8}{3}\arctan \frac{1}{2}x + C$ .