"I am Back" says Integration Part 9

We have $\begin{array}{rcl} \displaystyle I \; = \; \int_0^\infty \frac{\sin x}{x^2}(1 - e^{-x})\,dx & = & \displaystyle \int_0^\infty \frac{\sin x}{x} \int_0^1 e^{-xu}\,du\,dx \\ & = & \displaystyle \int_0^1 \int_0^\infty \frac{\sin x}{x} e^{-xu}\,dx\,du \end{array}$ If we define $F(u) \; = \; \int_0^\infty \frac{\sin x}{x} e^{-xu}\,du \;, \qquad \qquad u >0 \;,$ then $F'(u) \; = \; -\int_0^\infty \sin x e^{-xu}\,dx \; = \; -\frac{1}{u^2+1} \;, \qquad u > 0 \;,$ and hence $F(u) \; = \; \tfrac12\pi - \tan^{-1}u \;, \qquad u >0 \;,$ since $F(u)$ converges to $0$ as $u \to \infty$ . It is worth observing in passing that the fact that $F(u)$ tends to $\tfrac12\pi$ as $u \to 0$ evaluates the infinite integral $\int_0^\infty \frac{\sin x}{x}\,dx \; = \; \tfrac12\pi \;.$ Getting back on track, we see that $\begin{array}{rcl} I & = & \displaystyle \int_0^1 \left(\tfrac12\pi - \tan^{-1}u\right)\,du \; = \; \tfrac12\pi - \Big[u \tan^{-1}u\Big]_0^1 + \int_0^1 \frac{u}{u^2+1}\,du \\ & = & \displaystyle \tfrac12\pi - \tfrac14\pi + \Big[\tfrac12\ln(u^2+1)\Big]_0^1 \; = \; \tfrac14\pi + \tfrac12\ln2 \;, \end{array}$ so that $A=4$ and $B=2$ , and the answer is $\boxed{6}$ .