Evaluate, $I = \int\limits_0^\infty {\frac{{dx}}{{{{(1 + {x^2})}^5}}}} $
$Let,x = \tan \theta $
$\therefore dx = {\sec ^2}\theta d\theta $
$\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{(1 + {{\tan }^2}\theta )}^5}}}} d\theta $
$\therefore I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}\theta }}{{{{({{\sec }^2}\theta )}^5}}}} d\theta $
$\therefore I = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^8}\theta d\theta } $
$\therefore I = \frac{{7 \cdot 5 \cdot 3 \cdot 1}}{{8 \cdot 6 \cdot 4 \cdot 2}} \times \frac{\pi }{2} = \frac{{7 \cdot 5}}{{8 \cdot 2 \cdot 4 \cdot 2}} \times \frac{{22}}{{7.2}}$
$\therefore I = \frac{{55}}{{128}}$
