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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 :)

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  • $\begingroup$ Yes, My bad, I am new to LATEX formatting. $\endgroup$ – Ashish Gupta Mar 21 '18 at 16:47
  • $\begingroup$ 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. $\endgroup$ – Winther Mar 21 '18 at 16:53
  • $\begingroup$ 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. $\endgroup$ – Ashish Gupta Mar 21 '18 at 16:55
  • $\begingroup$ Winther, I am not getting you there. $\endgroup$ – Ashish Gupta Mar 21 '18 at 16:59
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    $\begingroup$ 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: $\frac{1}{2}$. $\endgroup$ – Winther Mar 21 '18 at 17:01
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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.

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  • $\begingroup$ Oh Ohk, Finally got it. Thank you for helping out, have a great day sir! $\endgroup$ – Ashish Gupta Mar 21 '18 at 17:09
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Hint:

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

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