对式子的等价转换除了有先加后减，还有先开方再平方

一、题目

$$I={\int }_2^{+\mathrm{\infty }}\frac{1}{x\sqrt{x-1}}\mathrm{d}x=?$$

难度评级：

二、解析

$$\begin{array}{rl}I& \\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{x\sqrt{x-1}}\mathrm{d}x\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{x}\cdot \frac{1}{\sqrt{x-1}}\mathrm{d}x\\ \\ & \Rightarrow (\sqrt{x-1}{)}^{\prime }=[(x-1{)}^{\frac{1}{2}}{]}^{\prime }=\frac{1}{2}\frac{1}{\sqrt{x-1}}\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{x}\cdot 2\mathrm{d}(\sqrt{x-1})\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{x+1–1}\cdot 2\mathrm{d}(\sqrt{x-1})\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{1+x–1}\cdot 2\mathrm{d}(\sqrt{x-1})\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{1+[(x–1{)}^{\frac{1}{2}}{]}^2}\cdot 2\mathrm{d}(\sqrt{x-1})\\ \\ & ={\int }_2^{+\mathrm{\infty }}\frac{1}{1+(\sqrt{x-1}{)}^2}\cdot 2\mathrm{d}(\sqrt{x-1})\\ \\ & \Rightarrow \{\begin{array}{l}t=\sqrt{x-1}\\ t\in (1,+\mathrm{\infty })\end{array}\\ \\ & =2{\int }_1^{+\mathrm{\infty }}\frac{1}{1+t^2}\mathrm{d}t\\ \\ & =2\arctan t{|}_1^{+\mathrm{\infty }}\\ \\ & =2\left(\frac{\pi }{2}–\frac{\pi }{4}\right)\\ \\ & =\frac{\mathit{\pi }}{\mathbf{2}}\end{array}$$

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