📝 题目
例 11 求不定积分 $\displaystyle{\int \frac{1}{1 + {x}^{3}}\mathrm{\;d}x}$ .
💡 答案与解析
解 注意到
$$ \int \frac{1}{1 + {x}^{3}}\mathrm{\;d}x = \int \frac{1 + x - x}{1 + {x}^{3}}\mathrm{\;d}x $$
$$ = \int \frac{1}{1 - x + {x}^{2}}\mathrm{\;d}x - \int \frac{x}{1 + {x}^{3}}\mathrm{\;d}x. \tag{1.4} $$
设 $\displaystyle{I = \int \frac{1}{1 + {x}^{3}}\mathrm{\;d}x,J = \int \frac{x}{1 + {x}^{3}}\mathrm{\;d}x}$ ,由 (1.4) 式,则有
$$ I + J = \int \frac{1}{1 - x + {x}^{2}}\mathrm{\;d}x $$
$$ = \frac{2}{3}\sqrt{3}\arctan \frac{2}{3}\sqrt{3}\left( {x - \frac{1}{2}}\right) + C, $$
$$ I - J = \int \frac{1 - x}{1 + {x}^{3}}\mathrm{\;d}x = \int \frac{1 - x + {x}^{2} - {x}^{2}}{1 + {x}^{3}}\mathrm{\;d}x $$
$$ = \int \frac{1}{1 + x}\mathrm{\;d}x - \int \frac{{x}^{2}}{1 + {x}^{3}}\mathrm{\;d}x $$
$$ = \ln \left( {x + 1}\right) - \frac{1}{3}\ln \left( {{x}^{3} + 1}\right) + C, $$
由此解得
$$ \int \frac{1}{1 + {x}^{3}}\mathrm{\;d}x = I = \frac{1}{3}\ln \left( {x + 1}\right) - \frac{1}{6}\ln \left( {{\left( x - \frac{1}{2}\right) }^{2} + \frac{3}{4}}\right) $$
$$ + \frac{1}{3}\sqrt{3}\arctan \frac{2}{3}\sqrt{3}\left( {x - \frac{1}{2}}\right) + C. $$
评注 本题虽然只要求计算单个积分 $I$ ,但是在计算过程中又 “冒出”一个新积分 $J$ ,将 $I,J$ 联立计算,要比单独计算 $I$ 简单得多.
\subsubsection{四、分部积分法}