📝 题目
例 22 问下列积分是否可积(即原函数是初等函数):
(1) $\displaystyle{\int \sqrt{1 + \frac{1}{x}}\mathrm{\;d}x}$ ; (2) $\displaystyle{\int \frac{\mathrm{d}x}{\sqrt{1 + {\sin }^{2}x}}}$ .
💡 答案与解析
解 (1) 原式 $= \int {x}^{-\frac{1}{2}}{\left( 1 + {x}^{2}\right) }^{\frac{1}{2}}\mathrm{\;d}x$ ,由此可见,
$$ p = \frac{1}{2},m = - \frac{1}{2},n = 2 \Rightarrow \frac{m + 1}{n} = \frac{1}{4},p + \frac{m + 1}{n} = \frac{3}{4}. $$
由于 $p,\frac{m + 1}{n},p + \frac{m + 1}{n}$ 三个量都非整数,从而原式不可积. (2)原式 $\displaystyle{= \int \frac{\cos x}{\sqrt{1 - {\sin }^{4}x}} = \int \frac{\mathrm{d}\sin x}{\sqrt{1 - {\sin }^{4}x}}}$
$$ \overset{t = \sin x}{ = }\int {\left( 1 - {t}^{4}\right) }^{-\frac{1}{2}}\mathrm{\;d}x, $$
由此可见
$$ p = - \frac{1}{2},m = 0,n = 4 \Rightarrow \frac{m + 1}{n} = \frac{1}{4},p + \frac{m + 1}{n} = - \frac{1}{4}. $$
由于 $p,\frac{m + 1}{n},p + \frac{m + 1}{n}$ 三个量都非整数,从而原式不可积.