考研数学不定积分补充例题 - 第4页共8页

题目 04

$I = \int \frac{d x}{x^{2} \sqrt{2 x - 4}} = ?$

本题主要解题思路为：整体代换去根号、三角代换。

$t = \sqrt{2 x - 4} \Rightarrow t^{2} = 2 x - 4 \Rightarrow x = \frac{t^{2} + 4}{2}$

$d x = t d t \Rightarrow$

$I = \int \frac{1}{x^{2} \sqrt{2 x - 4}} d x = \int \frac{1}{\frac{{(t^{2} + 4)}^{2}}{4} \cdot t} t d \Rightarrow$

$I = \int \frac{4}{{(t^{2} + 4)}^{2}} d t \Rightarrow$

$I = 4 \int \frac{1}{{(t^{2} + 4)}^{2}} d t \Rightarrow$

$I = 4 \times \frac{1}{16} \arctan (\frac{t}{2}) + 4 \times \frac{t}{8 (4 + t^{2})} + C \Rightarrow$

$I = \frac{1}{4} \arctan (\frac{t}{2}) + \frac{t}{2 (4 + t^{2})} + c$

又：

$t = \sqrt{2 x - 4} \Rightarrow$

因此：

$I = \frac{1}{4} \arctan (\frac{\sqrt{2 x - 4}}{2}) + \frac{\sqrt{2 x - 4}}{2 (2 x)} + C \Rightarrow$

$I = \frac{1}{4} \arctan (\frac{\sqrt{2 x - 4}}{2}) + \frac{\sqrt{2 x - 4}}{4 x} + C$

页码： 12345678