📝 题目
例 7 求 ${J}_{n} = \int \frac{\mathrm{d}x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}}$ .
💡 答案与解析
解 利用分部积分法得
$$ {J}_{n} = \frac{x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} - \int x\mathrm{\;d}\frac{1}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} $$
$$ = \frac{x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} + {2n}\int \frac{{x}^{2}}{{\left( {x}^{2} + {a}^{2}\right) }^{n + 1}}\mathrm{\;d}x $$
$$ = \frac{x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} + {2n}\int \frac{{x}^{2} + {a}^{2} - {a}^{2}}{{\left( {x}^{2} + {a}^{2}\right) }^{n + 1}}\mathrm{\;d}x $$
$$ = \frac{x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} + {2n}{J}_{n} - {2n}{a}^{2}{J}_{n + 1}. $$
由此得到递推公式
$$ {J}_{n + 1} = \frac{1}{{2n}{a}^{2}}\frac{x}{{\left( {x}^{2} + {a}^{2}\right) }^{n}} + \frac{{2n} - 1}{{2n}{a}^{2}}{J}_{n}. $$
因为我们已经知道
$$ {J}_{1} = \int \frac{\mathrm{d}x}{{x}^{2} + {a}^{2}} = \frac{1}{a}\arctan \frac{x}{a} + C, $$
所以利用上面的递推公式可以求得任何 ${J}_{n}$ .