$\int {\sqrt {\tan x} } dx$

Let $I = \int {\sqrt {\tan x} } dx$

Let's put $\tan x = {t^2} \Rightarrow {\sec ^2}xdx = 2tdt$

therefore,$I = \int t \cdot \frac{{2t}}{{{{\sec }^2}x}}dt = 2\int {\frac{{{t^2}}}{{1 + {t^4}}}} dt$

$= \int {\frac{{\left( {{t^2} + 1} \right) + \left( {{t^2} - 1} \right)}}{{\left( {1 + {t^4}} \right)}}} dt$

$= \int {\frac{{{t^2} + 1}}{{1 + {t^4}}}} dt + \int {\frac{{{t^2} - 1}}{{1 + {t^4}}}} dt$

$= \int {\frac{{1 + \frac{1}{{{t^2}}}}}{{{t^2} + \frac{1}{{{t^2}}}}}} dt + \int {\frac{{1 - \frac{1}{{{t^2}}}}}{{{t^2} + \frac{1}{{{t^2}}}}}} dt$

$= \int {\frac{{1 - \left( { - \frac{1}{{{t^2}}}} \right)dt}}{{{{\left( {t - \frac{1}{t}} \right)}^2} + 2}}} + \int {\frac{{1 + \left( { - \frac{1}{{{t^2}}}} \right)}}{{{{\left( {t + \frac{1}{t}} \right)}^2} - 2}}} dt$

Let's put $u = t - \frac{1}{t} \Rightarrow du = \left( {1 + \frac{1}{{{t^2}}}} \right)dt$

and $v = t + \frac{1}{t} \Rightarrow dv = \left( {1 - \frac{1}{{{t^2}}}} \right)dt$

therefore,$I = \int {\frac{{du}}{{{u^2} + {{(\sqrt 2 )}^2}}}} + \int {\frac{{dv}}{{{v^2} - {{(\sqrt 2 )}^2}}}}$

$= \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{u}{{\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{v - \sqrt 2 }}{{v + \sqrt 2 }}} \right| + C$

$= \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{{\tan x - 1}}{{\sqrt {2\tan x} }}} \right) + \frac{1}{{2\sqrt 2 }}\log \left| {\frac{{\tan x - \sqrt {2\tan x} + 1}}{{\tan x + \sqrt {2\tan x} + 1}}} \right| + C$