$$
\begin{aligned}
& I=\int_0^{\pi / 4} \sqrt{\tan x}+\sqrt{\cot x} d x \\
&=\int_0^{\pi / 4}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) d x \\
&=\int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}} d x \\
&=\int_0^{\pi / 4} \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2 \sin x \cos x}} d x \\
&=\sqrt{2} \int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} d x
\end{aligned}
$$
Let $\quad \sin x-\cos x=t$
$$
\Rightarrow \quad(\cos x+\sin x) d x=d t
$$
$$
\begin{aligned}
\therefore I & =\sqrt{2} \int_0^{\pi / 4} \frac{d t}{\sqrt{1-t^2}} \\
& =\sqrt{2}\left[\sin ^1 t\right]_0^{\pi / 4} \\
& =\sqrt{2}\left[\sin ^1(\sin x-\cos x)\right]_0^{\pi / 4} \\
& =\sqrt{2}\left[\sin ^1\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)-\sin ^1\left(\sin \infty^{\circ}-\cos \infty^{\circ}\right)\right] \\
& =\sqrt{2}\left[\sin ^1 0-\sin ^1(-1)\right] \\
& =\sqrt{2}\left[0+\frac{\pi}{2}\right]=\sqrt{2} \frac{\pi}{2}=\frac{\pi}{\sqrt{2}}
\end{aligned}
$$