A calculus problem by Rishabh Deep Singh

$\begin{aligned} I & = \lim_{m \to \infty} \int_{-\infty}^\infty \frac {dx}{1+x^2+x^4+...+x^{2m}} & \small {\color{#3D99F6}\text{Since the integral is even and absolutely convergent (see Note)}} \\ & = \lim_{m \to \infty} {\color{#3D99F6}2} \int_{\color{#3D99F6}0}^\infty \frac {dx}{1+x^2+x^4+...+x^{2m}} & \small {\color{#3D99F6}\text{Note that when }|x| \ge 1 \text{, the integrand}=0} \\ & = \lim_{m \to \infty} 2 \int_0^{\color{#3D99F6}1} \frac {dx}{1+x^2+x^4+...+x^{2m}} \\ & = 2 \int_0^1 (1-x^2) \ dx \\ & = 2 \left[x - \frac {x^3}3 \right]_0^1 \\ & = \frac 43 \end{aligned}$

$\implies A+B = 4+3 = \boxed{7}$

Note: Since $\displaystyle \int_{-\infty}^\infty \left|\frac 1{1+x^2+x^4+...+x^{2m}} \right| \ dx < \int_{-\infty}^\infty \frac 1{1+x^2} \ dx < \infty$ the integral is absolutely convergent and we can interchange integration with summation.