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Question Number 2770 by Yozzi last updated on 26/Nov/15

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

Answered by prakash jain last updated on 27/Nov/15

$$x-1={\tan }^{2}u$$$$dx=2\tan u{\sec }^{2}udu$$$$x={\sec }^{2}u$$$$\int \frac{dx}{x\sqrt{x-1}}=\int \frac{2\tan u{\sec }^{2}u}{{\sec }^{2}u\tan u}du=\int 2du=2u$$$$={2\tan }^{-1}\left(\sqrt{x-1}\right)$$$$limits$$$$=2[\frac{\pi }{2}-\frac{\pi }{4}]=\frac{\pi }{2}$$

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