Praise the king

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac 1{1+\tan^r x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^r x} + \frac 1{1+\tan^r \left(\frac \pi 2-x\right)} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^r x} + \frac 1{1+\cot^r x} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^r x} + \frac 1{1+\frac 1{\tan^r x}} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\tan^r x} + \frac {\tan^r x}{\tan^r x +1} \right) dx \\ & = \frac 12 \int_0^\frac \pi 2 dx = \frac \pi 4 \end{aligned}$

Therefore, $A+B = \frac 14 + 0 = \boxed{0.25}$ .

let $I(r) = \int_{0}^{\frac{1}{2}\pi} \dfrac{1}{1+\tan ^r x}\,dx=\int_{0}^{\frac{1}{2}\pi} \dfrac{\cos^r x }{\cos^r x +\sin^r x}\,dx$ Using the relation $\int_{a}^{b} f(x) \,dx=\int_{a}^{b} f(a+b-x)\,dx$ we can deduce the following $\int_{0}^{\frac{1}{2}\pi} \dfrac{\sin^r x }{\cos^r x +\sin^r x}\,dx$ we have $2I(r) =\int_{0}^{\frac{1}{2}\pi} \left(\dfrac{\cos^r x }{\cos^r x+\sin^r x}+\dfrac{\sin^r x }{\cos^r x +\sin^r x} \right)\,dx = \dfrac{\pi}{2}\implies I(r) =\dfrac{\pi}{4}$ so the required answer is $\frac{1}{4}+0=0.25$ .