Splitting Pairs

Putting $x = \tan\theta$ , the integral becomes $\begin{aligned} I & = \; \int_0^{\frac12\pi} \frac{3 + 2\cos\theta + 2\sin\theta}{(\sin\theta+1)^2(\cos\theta+1)^2}\,d\theta \; = \; \int_0^{\frac12\pi} \left(\frac{1}{(1 + \sin\theta)^2} + \frac{1}{(1 + \cos\theta)^2}\right)\,d\theta \\ & = \; 2\int_0^{\frac12\pi} \frac{d\theta}{(1 + \cos\theta)^2} \; = \; \tfrac12\int_0^{\frac12\pi} \sec^4\tfrac12\theta\,d\theta \\ & = \; \Big[\tan\tfrac12\theta + \tfrac13\tan^3\tfrac12\theta\Big]_0^{\frac12\pi} \; = \; \tfrac43 \end{aligned}$ making the answer $4+3 = \boxed{7}$ .

Let

$I=\int_0^{\infty }\dfrac{3(x^2+1)+2(x+1)\sqrt{x^2+1}}{\left(x+(x+1)\sqrt{x^2+1}+x^2+1\right)^2}\; dx.$

Observe that

$x+(x+1)\sqrt{x^2+1}+x^2+1=\left(1+\sqrt{x^2+1}\right)\left(x+\sqrt{x^2+1}\right)$

and

$3(x^2+1)+2(x+1)\sqrt{x^2+1}=\left(1+\sqrt{x^2+1}\right)^2+\left(x+\sqrt{x^2+1}\right)^2.$

Hence, $I$ can be split so that

$I=\int_0^{\infty }\dfrac{1}{\left(x+\sqrt{x^2+1}\right)^2}+\dfrac{1}{\left(1+\sqrt{x^2+1}\right)^2}\; dx = \overset{=X}{\overbrace{\int_0^{\infty }\dfrac{1}{\left(x+\sqrt{x^2+1}\right)^2}\; dx}} + \overset{=Y}{\overbrace{\int_0^{\infty }\dfrac{1}{\left(1+\sqrt{x^2+1}\right)^2}\; dx}}.$

Substituting $x=\frac1u$ gives $\begin{aligned} X&=\int_{\infty }^0\dfrac{1}{\left(\frac{1}{u}+\sqrt{\left(\frac1u\right)^2+1}\right)^2}\times \dfrac{-1}{u^2}\; du\\ &=\int_{\infty }^0\dfrac{-u^2}{u^2\left(1+\sqrt{u^2+1}\right)^2}\; du\\ &=\int_0^{\infty }\dfrac{1}{\left(1+\sqrt{u^2+1}\right)^2}\; du\\ &=Y. \end{aligned}$ Then $I=2X$ . Substituting $x=\frac{1-t^2}{2t}$ gives $\begin{aligned} X&=\int_1^0\dfrac{1}{\left(\frac{1-t^2}{2t}+\frac{1-t^2}{2t}+t\right)^2}\times \left(\dfrac{-1}{2t^2}-\dfrac12\right)\; dt\\ &=\int_0^1\dfrac{1}{2}+\dfrac{t^2}{2}\; dt\\ &=\left[\dfrac{t}{2}+\dfrac{t^3}{6}\right]^1_0\\ &=\dfrac{2}{3}. \end{aligned}$ Therefore, $I=2\times \frac{2}{3}=\frac{4}{3}$ , and hence $A+B=4+3=\boxed{7}$ .