I can't seem to solve this integration problem, despite many attempts. The question goes like this:
$$\int_{0}^{1}\frac{\tan^{-1}\left(\frac{x}{x+1}\right)\,{\rm d}x}{\tan^{-1}\left(\frac{1+2x-2x^2}{2}\right)} $$ I have tried using the rule of replacing $x$ by $1-x$ in the hopes, the numerator and denominator might cancel, but no! Then I also tried adding multiple integrals obtained on the way of simplification, but I could not reach to a result. Please help me figure this out. Thank You :)
On
Let $f(x) = \frac{\tan^{-1}\left(\frac{x}{x+1}\right)}{\tan^{-1}\left(\frac{1+2x-2x^2}{2}\right)}$ be the integrand. Start by showing that
$$g(x) = f\left(\frac{1}{2}+x\right) - \frac{1}{2}$$
is an odd function, i.e. $g(x) = -g(-x)$. This follows by applying the addition formula for $\tan^{-1}(\cdot)$ to $g(x) + g(-x)$. Finally integrating $g$ over $[-1/2,1/2]$ and relating this to the integral of $f$ gives you the integral you are after.
Hint:
$$\tan^{-1} \left(\frac{x}{x+1} \right)+ \tan^{-1} \left(\frac{1-x}{2-x} \right) = \tan^{-1} \left(\frac{1+2x-2x^2}{2} \right)$$