📝 题目
例 21 求不定积分 $I = \int \frac{\mathrm{d}x}{\sqrt[3]{{\left( x + 1\right) }^{2}{\left( x - 1\right) }^{4}}}$ .
💡 答案与解析
解法 1 记 $h\left( x\right) \overset{\text{ 定义 }}{ = }\sqrt[3]{{\left( x + 1\right) }^{2}{\left( x - 1\right) }^{4}} = \left( {{x}^{2} - 1}\right) \sqrt[3]{\frac{x - 1}{x + 1}}$ ,并令 $t = \sqrt[3]{\frac{x - 1}{x + 1}}$ ,利用对数微分法,则有
$$ \frac{\mathrm{d}t}{t} = \frac{\mathrm{d}x}{3\left( {x - 1}\right) } - \frac{\mathrm{d}x}{3\left( {x + 1}\right) } \Rightarrow \frac{\mathrm{d}x}{{x}^{2} - 1} = \frac{3\mathrm{\;d}t}{2t}, $$
$$ I = \int \frac{\mathrm{d}x}{h\left( x\right) } = \int \frac{3\mathrm{\;d}t}{2{t}^{2}} = - \frac{3}{2t} + C = - \frac{3}{2}\sqrt[3]{\frac{x + 1}{x - 1}} + C. $$
解法 2 记 $h\left( x\right) = \sqrt[3]{{\left( x + 1\right) }^{2}{\left( x - 1\right) }^{4}} = {\left( x - 1\right) }^{2} \cdot \sqrt[3]{{\left( \frac{x + 1}{x - 1}\right) }^{2}}$ , 并令 $t = \sqrt[3]{\frac{x + 1}{x - 1}}$ ,则有
$$ {t}^{3} = 1 + \frac{2}{x - 1} \Rightarrow 3{t}^{2}\mathrm{\;d}t = - \frac{2\mathrm{\;d}x}{{\left( x - 1\right) }^{2}}, $$
从而
$$ I = \int \frac{\mathrm{d}x}{h\left( x\right) } = \int - \frac{3{t}^{2}\mathrm{\;d}t}{2{t}^{2}} = - \frac{3}{2}t + C = - \frac{3}{2}\sqrt[3]{\frac{x + 1}{x - 1}} + C. $$
解法 ${3h}\left( x\right) = \sqrt[3]{{\left( x + 1\right) }^{2}{\left( x - 1\right) }^{4}} = {\left( x - 1\right) }^{2} \cdot \sqrt[3]{{\left( 1 + \frac{2}{x - 1}\right) }^{2}}$ , 并令 $t = 1 + \frac{2}{x - 1}$ ,则有
$$ \mathrm{d}t = - \frac{2\mathrm{\;d}x}{{\left( x - 1\right) }^{2}}, $$
$$ I = \int \frac{\mathrm{d}x}{h\left( x\right) } = - \frac{1}{2}\int {t}^{-\frac{2}{3}}\mathrm{\;d}t = - \frac{3}{2}\sqrt[3]{t} + C, $$
即得
$$ I = - \frac{3}{2}{\left( \frac{x + 1}{x - 1}\right) }^{\frac{1}{3}} + C. $$