Putnam, A3 (2016) Let $f: \mathbb R\rightarrow \mathbb R$ satisfying \begin{equation*} f(x) + f\bigg(1-{1\over x}\bigg) = \arctan x\end{equation*}
for all $x\neq 0$. I want to evaluate \begin{equation*} \int_0^1 f(x) dx \end{equation*}
I know that $\tan\bigg(f(x) + f\bigg(1-{\cfrac 1x}\bigg)\bigg) = x$ and $-\pi/2 < f(x) + f\bigg(1-{1\over x}\bigg) < \pi/2$, but none of this seems useful yet. Since $x\neq 0$, the answer should be the solution to
\begin{equation*}
\lim_{n\rightarrow 0}\int_n^1f(x)dx\end{equation*}
But again, I'm still unsure where to go from here. Any hint would be appreciated.
 A: Answer being revised
Noting that on $]0,\pi/2[$ :
$$\arctan(\frac{1}{x})+\arctan(x)=\dfrac{\pi}{2} $$
Then your relation can be writen
$$ f(x)+f(1-1/x)=\arctan(x) $$
$$ f(1-1/x)+f(x) = \dfrac{\pi}{2}-\arctan(\frac{1}{x})=\dfrac{\pi}{2}-(f(1/x)+f(1-x))$$
First
\begin{align*}
 \int_0^1 f(1-1/x)+f(1/x)dx&  = \dfrac{\pi}{2}-\int_0^1 f(1-x)+f(x)dx   \\
\int_1^\infty \dfrac{f(1-u)+f(u)}{u^2}du&= \dfrac{\pi}{2}-2\int_0^1f(x)dx
\end{align*}
Second
Composing two times in addition, as suggested in comments by @Daniel Schepler $ x \to 1-1/x$ :
\begin{align*} f(\frac{1}{1-x}) + f(x) & =\arctan(\frac{1}{1-x})\\
&=\text{sgn}(1-x)\dfrac{\pi}{2}-\arctan(1-x)\\
& =\text{sgn}(1-x)\dfrac{\pi}{2}-(f(1-\frac{1}{1-x})+f(1-x))\\
\end{align*}
Taking $x \in ]0,1[$,
\begin{align*}\int_0^1 f(1-\frac{1}{1-x})+f(\frac{1}{1-x})dx&=\dfrac{\pi}{2}-\int_0^1 f(1-x)+f(x)dx \\
&= \dfrac{\pi}{2}-2\int_0^1f(x)dx \\
\end{align*}
So
$$ -\int_1^\infty \dfrac{f(1-u)+f(u)}{u^2}du=\dfrac{\pi}{2}-2\int_0^1f(x)dx $$
Conclusion by summing the two part recognizing opposite integrals, if everything is ok.

 $$ \int_0^1 f(x)dx=\dfrac{\pi}{4}$$

