Question

Evaluate the following : ${\displaystyle {\int }_1^{\infty }\frac{1}{\sqrt{x}(1+x)}\cdot dx}$

Answer 1

Let I = ${\displaystyle {\int }_1^{\infty }\frac{1}{\sqrt{x}(1+x)}\cdot dx}$

Put x = tan²t

∴ dx = ${\displaystyle \left[2\tan t\frac{d}{dt}\left(\tan t\right)\right]\cdot dt}$

= 2 tan t sec²t·dt

When x = ${\displaystyle \infty ,{\tan }^{2}t=\infty \therefore t=\frac{\pi }{2}}$

When x = ${\displaystyle 1, {\tan }^{2}t=1 \therefore t=\frac{\pi }{4}}$

∴ I = ${\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{2\tan t{\sec }^{2}t}{\sqrt{{\tan }^{2}t}(1+{\tan }^{2}t)}\cdot dt}$

= ${\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{2{\sec }^{2}t}{{\sec }^{2}t}\cdot dt}$

= ${\displaystyle 2{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}1\cdot dt=2{\left[t\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}}$

= ${\displaystyle 2[\frac{\pi }{2}-\frac{\pi }{4}]}$

= ${\displaystyle 2\left[\frac{\pi }{4}\right]}$

= ${\displaystyle \frac{\pi }{2}}$.