How to prove a fomula is real or not? The formula is:
$$f(u):=\sqrt{1+i}\arctan\left(\dfrac{\sqrt{2}}{\sqrt{(-1+i)u}}\right)+\sqrt{1-i}\arctan\left(\frac{(-1)^{3/4}\sqrt{(-1-i)u}}{u}\right)\tag{1}$$
I obtained it from the following definite integral:
$$f(u):=\int_u^{\infty } \frac{\sqrt{x}}{x^2-2 x+2} \, dx\quad {\text{ where }} u>0\tag{2}$$
Numercial estimation indicates that, when $u>0$, $f(u)\in \mathbb{R}$ because the imaginary part of the results are close to numerical epsilon.
How to prove or disprove it? If the conclusion is true, is it possible to eliminate the imaginary unit $i$ from the fomula of $f(u)$ in $(1)$ and obtain a relatively simpler form of it which does not contain or implies any imaginary number?
update
$$f^*(u):=\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} \left(\left(\sqrt{2}-1\right) \left(\ln \left(u-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)-\ln \left(u+\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)\right)-2 \arctan\left(-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)+2 \arctan\left(\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)-2 \pi \right)$$

Inspired by How to simplify $\Re\left[\sqrt2 \tan^{-1} {x\over \sqrt i}\right]$?
I manage to write a primitive of $\frac{\sqrt{x}}{x^2-2 x+2}$ using real functions. Let
$$F(x):=\frac{1}{\sqrt{2\sqrt{2}-2}}\cdot 
\arctan\left(\frac{\sqrt{2\sqrt{2}-2}\cdot \sqrt{x}}{x-\sqrt{2}}\right)$$
$$G(x):=\frac{1}{\sqrt{2\sqrt{2}+2}}\cdot 
\mbox{arctanh}\left(\frac{\sqrt{2\sqrt{2}+2}\cdot\sqrt{x}}{x+\sqrt{2}}\right),$$
then you can verify that
$$\frac{d}{dx}\left(-F(x)-G(x)\right)= \frac{\sqrt{x}}{x^2-2 x+2}.$$
