Explanation
Let,
$$ I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{dx}{1+\sqrt{\tan x}}} $$
$$ I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{dx}{1+\sqrt{\dfrac{\sin x}{\cos x}}}} $$
$$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}$$ …… (1)
We know that,
$$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$$
Therefore,
$$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}}{\sqrt{\cos \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}+\sqrt{\sin \left( \dfrac{\pi }{6}+\dfrac{\pi }{3}-x \right)}}dx}$$
$$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}dx}$$
$$I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$$ …… (2)
Add equations (1) and (2).
$$ 2I=\int\limits_{\pi /6}^{\pi /3}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx} $$
$$ 2I=\int\limits_{\pi /6}^{\pi /3}{1\cdot dx} $$
$$ 2I=\left[ x \right]_{\pi /6}^{\pi /3} $$
$$ 2I=\dfrac{\pi }{3}-\dfrac{\pi }{6} $$
$$ 2I=\dfrac{\pi }{6} $$
$$ I=\dfrac{\pi }{12} $$
Hence, this is the required value of the integral.
$$\int _{ 1 }^{ x }{ \cfrac { dt }{ \left| t \right| \sqrt { { \left| t \right| }^{ 2 }-1 } } } =\cfrac { \pi }{ 6 } \\ put\quad \left| t \right| =\sec { \theta } \Rightarrow dt\sec { \theta } \tan { \theta } d\theta \\ changing\quad limit,\\ t=x\Rightarrow \sec { \theta } =x\Rightarrow \sec ^{ -1 }{ x } \\ t=1\Rightarrow \sec { \theta } =1\Rightarrow \sec ^{ -1 }{ 1 } \\ \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \cfrac { \sec { \theta \tan { \theta } d\theta } }{ \sec { \theta } \sqrt { \sec ^{ 2 }{ \theta } -1 } } =\cfrac { \pi }{ 6 } } \\ (\because \sec ^{ 2 }{ \theta } -1=\tan ^{ 2 }{ \theta } )\\ \Rightarrow \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \cfrac { \sec { \theta \tan { \theta } d\theta } }{ \sec { \theta } \tan { \theta } } =\cfrac { \pi }{ 6 } } \\ \Rightarrow \int _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ d\theta =\cfrac { \pi }{ 6 } } \Rightarrow _{ \sec ^{ -1 }{ 1 } }^{ \sec ^{ -1 }{ x } }{ \left[ \theta \right] }=\cfrac { \pi }{ 6 } \\ \therefore \sec ^{ -1 }{ x } -\sec ^{ -1 }{ 1 } =\cfrac { \pi }{ 6 } \\ \therefore \sec ^{ -1 }{ x } =\cfrac { \pi }{ 6 } +\sec ^{ -1 }{ 1 } =\cfrac { \pi }{ 6 } +0\\ \therefore x=\sec { \cfrac { \pi }{ 6 } } =\cfrac { 2 }{ \sqrt { 3 } } $$
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