2-dx-x-x-1-

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

\int_{2}^{\infty} \frac{d x}{x \sqrt{x - 1}} = ?

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

x - 1 = \tan^{2} u

d x = 2 \tan u \sec^{2} u d u

x = \sec^{2} u

\int \frac{d x}{x \sqrt{x - 1}} = \int \frac{2 \tan u \sec^{2} u}{\sec^{2} u \tan u} d u = \int 2 d u = 2 u

= {2 \tan}^{- 1} (\sqrt{x - 1})

l i m i t s

= 2 [\frac{π}{2} - \frac{π}{4}] = \frac{π}{2}