Find the integral $\int_{0}^{1} f(x)dx$ for $f(x)+f(1-{1\over x})=\arctan x\,,\quad \forall \,x\neq 0$. Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$$f(x)+f\left(1-{1\over x}\right)=\arctan x\,,\quad \forall \,x\neq 0$$
Find $$\int_{0}^1 f(x)\,dx$$
My Attempt :
Replace $x$ by $1/x$ in given equation
$$f\left({1\over x}\right)+f(1-x)=\arctan {1\over x}$$Add both equations 
$$f(x)+f\left(1-{1\over x}\right)+f\left({1\over x}\right)+f(1-{x})=\arctan x\,+\arctan {1\over x}$$Rearranging thenm gives
$$f(x)+f(1-x)+f\left({1\over x}\right)+f\left(1-{1\over x}\right)={\pi\over2}$$Now it seems to me that $f(x)=f\left({1\over x}\right)$ Am I correct here? (I don't have proof though)
$$f(x)+f(1-x)={\pi\over 4}$$ $$\int_0^1 f(x)\,dx =\int_0^1f(1-x)\, dx={\pi\over 8}$$
I'm not sure about my assumption. Thank you 
 A: Appearently $f(x)=f(1/x)$ does not hold.

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 $$\int_0^1 f(x) dx = \frac{3\pi}{8}$$
A: 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.
A: An Attempt: Let
\begin{equation}
f(x) + f(1 - \frac{1}{x}) = arctan(x)   \tag{1}
\end{equation}
Integrating:
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \int_0^1 arctan(x) dx 
\end{equation}
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \frac{1}{4}(\pi-\ln 4) \tag{2}
\end{equation}
Now using one of your results:
$$f(\frac{1}{x}) + f(1 - x) = arctan(\frac{1}{x})$$
and integrating and plugging in value for the RHS
\begin{equation}
\int_0^1 f(\frac{1}{x}) dx + \int_0^1 f(1 - x) dx = \frac{1}{4}(\pi+\ln 4) \tag{3}
\end{equation}
Now using another of your results:
$$ f(x) + f(1 - \frac{1}{x})  + f(\frac{1}{x}) + f(1 - x)  = \frac{\pi}{2}$$
Integrating:
$$
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx +\int_0^1 f(\frac{1}{x}) dx + \int_0^1 f(1 - x) dx =\frac{\pi}{2}$$
Using the fact that
$$\int_0^1 f(x) dx  = \int_0^1 f(1 - x) dx \tag{a}$$ 
we have:
$$
2\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx +\int_0^1 f(\frac{1}{x}) dx  =\frac{\pi}{2}
$$
Now substituting the value of $\displaystyle \int_0^1 f(\frac{1}{x}) dx $ from (3) into above result and using (a):
$$
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx =\frac{\pi}{2} - \frac{1}{4}(\pi+\ln 4) \tag{4}
$$
Leads no where. General procedure is fine but didn't work. 
