\(D=\frac{4x+3}{x^2+1}\Leftrightarrow D\left(x^2+1\right)=4x+3\)
\(\Leftrightarrow Dx^2+D-4x-3=0\)
\(\Leftrightarrow Dx^2-4x+\left(D-3\right)=0\)
\(\Delta'=4-D\left(D-3\right)\ge0\Rightarrow4-D^2+3D\ge0\)
\(\Rightarrow\left(4-D\right)\left(D+1\right)\ge0\Rightarrow-1\le D\le4\)