$\int_0^{1/2} {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}}$

Let $I = \int_0^{1/2} {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}}$

Let's put $x = \sin \theta$

$\Rightarrow$ $dx = \cos \theta d\theta$
As $x \to 0,$ then $\theta \to 0$

and $x \to \frac{1}{2},$ then $\theta \to \frac{\pi }{6}$

therefore,$I = \int_0^{\pi /6} {\frac{{\cos \theta }}{{\left( {1 + {{\sin }^2}\theta } \right)\cos \theta }}} d\theta = \int_0^{\pi /6} {\frac{1}{{1 + {{\sin }^2}\theta }}} d\theta$

$= \int_0^{\pi /6} {\frac{1}{{{{\cos }^2}\theta \left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right)}}} d\theta$

$= \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta + {{\tan }^2}\theta }}} d\theta$

$= \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{1 + {{\tan }^2}\theta + {{\tan }^2}\theta }}} d\theta$

$= \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{1 + 2{{\tan }^2}\theta }}} d\theta$

Again, Let's put $\tan \theta = t$

$\Rightarrow$ ${\sec ^2}\theta d\theta = dt$
As $\theta \to 0,$ then $t \to 0$

and $\theta \to \frac{\pi }{6}$, then $t \to \frac{1}{{\sqrt 3 }}$

therefore,$I = \int_0^{1/\sqrt 3 } {\frac{{dt}}{{1 + 2{t^2}}}} = \frac{1}{2}\int_0^{1/\sqrt 3 } {\frac{{dt}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {t^2}}}}$

$= \frac{1}{2} \cdot \frac{1}{{1/\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right]_0^{1/\sqrt 3 } = \frac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ - 1}}(\sqrt 2 t)} \right]_0^{1/\sqrt 3 }$

$= \frac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\sqrt {\frac{2}{3}} - 0} \right] = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt {\frac{2}{3}} } \right)$