1708\hspace{-16}$Let $\bf{\mathbb{I}=\int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx = \int\frac{\tan x+1}{\sqrt{\tan x}}dx}$\\\\\\ Now Let $\bf{\tan x= t^2\,}$ Then $\bf{\sec^2 xdx = 2tdt}$\\\\\\ So $\bf{dx = \frac{2t}{\sec^2 x}dt = \frac{2t}{1+t^4}dt}$\\\\\\ So Integral $\bf{\mathbb{I} = \int\frac{t^2+1}{t}\cdot \frac{2t}{1+t^4}dt}$\\\\\\ So $\bf{\mathbb{I} = 2\int\frac{1+t^2}{1+t^4}dt = 2\int\frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+\left(\sqrt{2}\right)^2}dt}$\\\\\\ Now Let $\bf{\left(t-\frac{1}{t}\right)=u\;,}$ Then $\bf{\left(1+\frac{1}{t^2}\right)dt = du}$\\\\\\ So $\bf{\mathbb{I} = 2\int\frac{1}{t^2+\left(\sqrt{2}\right)^2}dt = \sqrt{2}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+\mathbb{C}}$\\\\\\ So $\bf{\mathbb{I} =\int \left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx = \sqrt{2}\tan^{-1}\left(\frac{\tan x-1}{\sqrt{2\cdot \tan x}}\right)+\mathbb{C}}$
