同济高数 第4章 第4-5-21题

[AI解答]

[AI解答] 我们要求解不定积分 $$ \int \frac{dx}{x^{2} \sqrt{2x-1}}. $$

**第一步：变量代换** 令 $$ t = \sqrt{2x-1}, \quad 则 \quad t^{2} = 2x - 1 \quad\Rightarrow\quad x = \frac{t^{2}+1}{2}. $$ 对 $x$ 求微分： $$ dx = t\, dt. $$

**第二步：代入积分** 原积分变为 $$ \int \frac{1}{x^{2} \sqrt{2x-1}}\,dx = \int \frac{1}{\left(\frac{t^{2}+1}{2}\right)^{2} \cdot t} \cdot t\, dt = \int \frac{1}{\frac{(t^{2}+1)^{2}}{4}} \, dt = 4 \int \frac{dt}{(t^{2}+1)^{2}}. $$

**第三步：计算三角函数积分** 对于 $$ \int \frac{dt}{(t^{2}+1)^{2}}, $$ 令 $t = \tan \theta$，则 $dt = \sec^{2}\theta\, d\theta$，且 $t^{2}+1 = \sec^{2}\theta$，于是 $$ \int \frac{dt}{(t^{2}+1)^{2}} = \int \frac{\sec^{2}\theta}{\sec^{4}\theta}\, d\theta = \int \cos^{2}\theta\, d\theta. $$

**第四步：计算 $\cos^{2}\theta$ 的积分** 利用公式 $\cos^{2}\theta = \frac{1+\cos 2\theta}{2}$，得 $$ \int \cos^{2}\theta\, d\theta = \frac{1}{2}\int (1+\cos 2\theta)\, d\theta = \frac{1}{2}\left( \theta + \frac{1}{2}\sin 2\theta \right) + C. $$ 又因为 $\sin 2\theta = 2\sin\theta\cos\theta$，而 $$ \sin\theta = \frac{t}{\sqrt{1+t^{2}}},\quad \cos\theta = \frac{1}{\sqrt{1+t^{2}}}, $$ 所以 $$ \frac{1}{2}\sin 2\theta = \sin\theta\cos\theta = \frac{t}{1+t^{2}}. $$ 因此 $$ \int \frac{dt}{(t^{2}+1)^{2}} = \frac{1}{2}\left( \arctan t + \frac{t}{1+t^{2}} \right) + C. $$

**第五步：回代** 原积分 $$ 4 \int \frac{dt}{(t^{2}+1)^{2}} = 2\left( \arctan t + \frac{t}{1+t^{2}} \right) + C. $$ 将 $t = \sqrt{2x-1}$ 代回： $$ \int \frac{dx}{x^{2}\sqrt{2x-1}} = 2\left( \arctan \sqrt{2x-1} + \frac{\sqrt{2x-1}}{1+(2x-1)} \right) + C. $$ 注意 $1+(2x-1) = 2x$，所以 $$ \int \frac{dx}{x^{2}\sqrt{2x-1}} = 2\arctan\sqrt{2x-1} + \frac{\sqrt{2x-1}}{x} + C. $$

**最终结果** $$ \boxed{2\arctan\sqrt{2x-1} + \frac{\sqrt{2x-1}}{x} + C}. $$

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