# How to solve this definite integration problem?

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 :)

• Yes, My bad, I am new to LATEX formatting. – Ashish Gupta Mar 21 '18 at 16:47
• Let $f(x)$ be the integrand then $g(x) = f(x + 1/2) - \frac{1}{2}$ is an odd function, i.e. $g(x) = -g(-x)$. Show this and then integrate $g$ over $[-1/2,1/2]$ to get the result. – Winther Mar 21 '18 at 16:53
• I replaced x by 1-x,( By rule of definite integration). Then I added two integrals. The denominators are the same! The numerators are different, so I added them up by using identities of tan-1. But could not derive a result. – Ashish Gupta Mar 21 '18 at 16:55
• Winther, I am not getting you there. – Ashish Gupta Mar 21 '18 at 16:59
• The point I'm trying to get at is that the integral of an odd function over $[-a,a]$ is zero and the integral of $g$ can be related to the integral of $f$. This gives a very simple result for the integral: $\frac{1}{2}$. – Winther Mar 21 '18 at 17:01

$$\tan^{-1} \left(\frac{x}{x+1} \right)+ \tan^{-1} \left(\frac{1-x}{2-x} \right) = \tan^{-1} \left(\frac{1+2x-2x^2}{2} \right)$$
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.