499 Followers Problem

Let

$\begin{aligned} I&= \displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}\dfrac{\cos^{-1}\left ( \dfrac{2x}{1+x^{2}} \right )+\tan^{-1}\left ( \dfrac{2x}{1-x^{2}} \right )}{e^{x}+1} \, \mathrm{d}x} \\&= \displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}\dfrac{\cos^{-1}\left ( \dfrac{2x}{1+x^{2}} \right )+\sin^{-1}\left ( \dfrac{2x}{1+x^{2}} \right )}{e^{x}+1} \, \mathrm{d}x} & \left( \small \color{#3D99F6} \tan^{-1}\left ( \dfrac{2x}{1-x^{2}} \right)=\sin^{-1} \left( \dfrac{2x}{1+x^{2}} \right); \right) \\&= \displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}\dfrac{\pi}{2\left( e^x+1\right)} \, \mathrm{d}x} & \left( \small \color{#3D99F6} \text{Using the identity } \sin^{-1}{a}+\cos^{-1}{a}= \dfrac{\pi}{2}; \text{ where } a \epsilon [-1,1] \right) \\&=\dfrac{1}{2}.\dfrac{\pi}{2}\displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}1. \, \mathrm{d}x} & \left( \small \color{#3D99F6} \displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}\dfrac{1}{\left( e^x+1\right)} \, \mathrm{d}x}=\dfrac{1}{2}\displaystyle{ \int_{{-1}/{\sqrt{3}}}^{{1}/{\sqrt{3}}}1.\mathrm{d}x} \right) \\&=\dfrac{\pi}{4}.\left[x \right]_{-1/\sqrt3}^{1/\sqrt3} \\&=\dfrac{2\pi}{4\sqrt3}=\boxed{\dfrac{\pi}{2\sqrt3}} \\& \boxed{A+B = 5} \end{aligned}$

Note: $\tan^{-1}\left ( \dfrac{2x}{1-x^{2}} \right)=\sin^{-1} \left( \dfrac{2x}{1+x^{2}} \right)$ ; because in a right triangle with sides $(1+x^2), (1-x^2)$ and $2x$ , the given relation holds true.