📝 题目
例 5 求 $\displaystyle{\int \sqrt{{x}^{2} - {a}^{2}}\mathrm{\;d}x}$ 和 $\displaystyle{\int \sqrt{{x}^{2} + {a}^{2}}\mathrm{\;d}x}$ .
💡 答案与解析
解 利用分部积分法得
$$ \int \sqrt{{x}^{2} - {a}^{2}}\mathrm{\;d}x = x\sqrt{{x}^{2} - {a}^{2}} - \int x\mathrm{\;d}\sqrt{{x}^{2} - {a}^{2}} $$
$$ = x\sqrt{{x}^{2} - {a}^{2}} - \int \frac{{x}^{2}}{\sqrt{{x}^{2} - {a}^{2}}}\mathrm{\;d}x $$
$$ = x\sqrt{{x}^{2} - {a}^{2}} - \int \frac{{a}^{2} + {x}^{2} - {a}^{2}}{\sqrt{{x}^{2} - {a}^{2}}}\mathrm{\;d}x $$
$$ = x\sqrt{{x}^{2} - {a}^{2}} - {a}^{2}\int \frac{\mathrm{d}x}{\sqrt{{x}^{2} - {a}^{2}}} - \int \sqrt{{x}^{2} - {a}^{2}}\mathrm{\;d}x. $$
由此得到
$$ \int \sqrt{{x}^{2} - {a}^{2}}\mathrm{\;d}x = \frac{x}{2}\sqrt{{x}^{2} - {a}^{2}} - \frac{{a}^{2}}{2}\int \frac{\mathrm{d}x}{\sqrt{{x}^{2} - {a}^{2}}} $$
$$ = \frac{x}{2}\sqrt{{x}^{2} - {a}^{2}} - \frac{{a}^{2}}{2}\ln \left| {x + \sqrt{{x}^{2} - {a}^{2}}}\right| + C. $$
用类似的办法可以求得
$$ \int \sqrt{{x}^{2} + {a}^{2}}\mathrm{\;d}x = \frac{x}{2}\sqrt{{x}^{2} + {a}^{2}} + \frac{{a}^{2}}{2}\ln \left| {x + \sqrt{{x}^{2} + {a}^{2}}}\right| + C. $$