Value of integral under certain condition If f is a real valued function and satisfies $f (x)+f (1-\frac {1}{x})=\arctan (x) $ for all $x $ except $0$ then find the least integer greater than or equal to $N=\int _0 ^1 f (x)dx $  . I  havent done any progress . What I tried was putting $x $  as $\frac {x-1}{x} $  to get $f(\frac {x-1}{x})+f (\frac {1}{1-x})=\arctan (\frac {x-1}{x}) $. I think we have to find some relation between like $f (x)+f (1-x)=c $ where c is some constant and use $N=\int _0 ^1 \frac {1}{2}c $ to get the required value.
 A: Using your hints, we have, substituting $x$ with $\frac{x-1}{x}$ and then with $\frac{1}{1-x}$, $$ f(x) + f\left( 1 - \frac{1}{x}\right) = \arctan(x)$$ $$ f\left( \frac{x-1}{x}\right) + f\left(\frac{1}{1-x} \right) = \arctan\left(\frac{x-1}{x}\right)$$ and $$ f(x) + f \left( \frac{1}{1-x} \right) = \arctan\left( \frac{1}{1-x}\right) $$
We would like to isolate $f(x)$. Hence we add the first and third equations and subtracting the second gives : $$ 2f(x) = \arctan(x) + \arctan\left( \frac{1}{1-x}\right) - \arctan\left( \frac{x-1}{x}\right) $$
Now we want to use the relation $\arctan(x) + \arctan\left( \frac{1}{x}\right) = \pi/2$. To do so just compute \begin{align*} 2(f(x) + f(1-x)) & = (\arctan(x) + \arctan(1/x))\\ & + \left(\arctan(1-x) + \arctan \left( \frac{1}{1-x}\right) \right) \\ & - \left(  \arctan\left( \frac{x-1}{x} \right) + \arctan \left( \frac{x}{x-1} \right)\right)\\ & = \frac{3\pi}{2} \end{align*}
Hence $$ 4 \int_0^1 f(x)dx = \int_0^1 2(f(x) + f(1-x))dx = \frac{3\pi}{2}$$ Finally $$ \int_0^1f(x)dx = \frac{3 \pi}{8}$$
