Show that $\int_0^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac12(\frac12+\frac1{\pi +2}-A)$

Just a couple of tiny missteps by the OP.

\begin{align*} \mathcal{I} \equiv\int_0^{\pi/2} \frac{ \sin x \cos x}{x+1}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{ \sin 2x }{2x+2}\,\mathrm{d}x \tag*{, then denote $u = 2x$} \\ &= \frac12\int_0^{\pi} \frac{ \color{blue}{\sin u} }{u+2}\,\color{blue}{\mathrm{d}u} \tag*{, then by-part ...} \end{align*} ... let $\color{blue}{\sin u \,\mathrm{d}u}$ go together \begin{align*} \mathcal{I} &= \frac12 \left[ \frac{ -\cos u }{u+2}\Bigg|_0^{\pi} - \int_0^{\pi} (-\cos u) \cdot \frac{ -1 }{ (u+2)^2 }\,\mathrm{d}u\right ] \\ &= \frac12 \left[ \frac1{ \pi + 2} - \frac{-1}{2} - \int_0^{\pi} \frac{ \cos u}{ (u+2)^2 }\,\mathrm{d}u\right] \end{align*} which is the desired result.

You are on the right lines!

$$\small\int_0^{\pi/2} {\frac{\sin2x}{2x+2}dx}=\left[-\frac{1}{2x+2}\cos2x\right]_0^{\pi/2}-\int_0^{\pi/2} {\frac{\cos2x}{(2x+2)^2}dx}=\frac12\left(\frac12+\frac1{\pi+2}\right)-\int_0^{\pi/2} {\frac{\cos2x}{(2x+2)^2}dx}$$ Now use the substitution $u=2x$.