Find the integral $\int_{0}^{1} f(x)dx$ for $f(x)+f(1-{1\over x})=\arctan x\,,\quad \forall \,x\neq 0$.

Appearently $f(x)=f(1/x)$ does not hold.

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Let $g(x) = 1-1/x$, then we have $$g^2 (x) = 1/(1-x) \quad \quad \color{red}{g^3 (x) = x}$$

Hence $f(x) + f(g(x)) = \arctan x$ implies $$f(g(x)) + f(g^2(x)) = \arctan(g(x))$$ $$f(g^2(x)) + f(x) = \arctan(g^2(x))$$

Solving for $f(x)$ from these three equations give $$f(x) = \frac{1}{2}\left[\arctan x - \arctan(1-\frac{1}{x}) + \arctan(\frac{1}{1-x})\right]$$

and routine integration gives $$\int_0^1 f(x) dx = \frac{3\pi}{8}$$

If you set $h(x)=1-\frac1x$ your functional equation becomes $$ f(x)+f(h(x)) = \arctan(x) $$ A bit of algebra also shows that $h(h(h(x)))=x$; this allows us to write down a system of equations $$ \begin{array}{crrl} f(x) & {}+ f(h(x)) && {}= \arctan(x) \\ & f(h(x)) & {}+ f(h(h(x))) & {}= \arctan(h(x)) \\ f(x) && {}+ f(h(h(x))) & {}= \arctan(h(h(x))) \end{array} $$ which you can solve to find an explicit formula for $f(x)$: $$ f(x) = \frac{\arctan(x) - \arctan(1-\frac1x) + \arctan(\frac{1}{1-x})}2 $$ This does not satisfy $f(x)=f(\frac1x)$ like you assumed.