If $\underset{x\to \infty }{\lim }4x(\frac{\pi }{4}-{\tan }^{-1}\frac{x+1}{x+2})=y^2+4y+5$, then $\text{'}y\text{'}$ can be equal to

$$\begin{gathered} \underset{x\to \infty }{\lim }4x(\frac{\pi }{4}-{\tan }^{-1}\frac{x+1}{x+2})=\underset{x\to \infty }{\lim }4x\left({\tan }^{-1}\frac{1-\frac{x+1}{x+2}}{1+\frac{x+1}{x+2}}\right)=y^2+4y+5 \\ \Rightarrow \underset{x\to \infty }{\lim }4x\left({\tan }^{-1}\frac{1}{2x+3}\right)=y^2+4y+5 \\ \Rightarrow \underset{x\to \infty }{\lim }4x\frac{{\tan }^{-1}\left(\frac{1}{2x+3}\right)}{\frac{1}{2x+3}}\cdot \left(\frac{1}{2x+3}\right)=y^2+4y+5 \\ \Rightarrow \underset{x\to \infty }{\lim }4\frac{{\tan }^{-1}\left(\frac{1}{2x+3}\right)}{\left(\frac{1}{2x+3}\right)}\cdot \frac{1}{2+\frac{3}{x}}=y^2+4y+5 \\ \Rightarrow 4\left(1\right)\frac{1}{(2+0)}=y^2+4y+5 \\ \Rightarrow 2=y^2+4y+5 \\ \Rightarrow y^2+4y+3=0\Rightarrow y^2+3y+y+3=0 \\ (y+3)(y+1)=0\Rightarrow y=-1 \left(or\right) -3 \end{gathered}$$