kaoyan2advanced 高等数学 第44题

**答案**：$\displaystyle \frac{10}{3}$ **解析**：步骤1：区间$[2,4]$，被积函数分母$\sqrt{|x^{2}-9|}$，在$x=3$处分段。步骤2：$\displaystyle \int_{2}^{4}\frac{x dx}{\sqrt{|x^{2}-9|}} = \int_{2}^{3}\frac{x dx}{\sqrt{9-x^{2}}} + \int_{3}^{4}\frac{x dx}{\sqrt{x^{2}-9}}$。步骤3：第一积分，令$u=9-x^{2}$，$du=-2x dx$，得$\displaystyle -\frac{1}{2}\int_{5}^{0}u^{-1/2}du = \frac{1}{2}\int_{0}^{5}u^{-1/2}du = \sqrt{u}\big|_{0}^{5}=\sqrt{5}$。第二积分，令$v=x^{2}-9$，$dv=2x dx$，得$\displaystyle \frac{1}{2}\int_{0}^{7}v^{-1/2}dv = \sqrt{v}\big|_{0}^{7}=\sqrt{7}$。步骤4：总和$\sqrt{5}+\sqrt{7}$，但答案$\displaystyle \frac{10}{3}$，数值近似$\sqrt{5}+\sqrt{7}\approx2.236+2.646=4.882$，$\displaystyle \frac{10}{3}\approx3.333$，不符。重新计算：$\displaystyle \int_{2}^{3}\frac{x}{\sqrt{9-x^{2}}}dx = -\sqrt{9-x^{2}}\big|_{2}^{3}=0-(-\sqrt{5})=\sqrt{5}$，$\displaystyle \int_{3}^{4}\frac{x}{\sqrt{x^{2}-9}}dx = \sqrt{x^{2}-9}\big|_{3}^{4}=\sqrt{7}-0=\sqrt{7}$，和为$\sqrt{5}+\sqrt{7}$，非$\displaystyle \frac{10}{3}$，可能答案有误。 **难度**：★★☆☆☆