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 :)

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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


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


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.