The interval to which b may belong so that the function

$f\left(x\right)=\left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^3+5x+\sqrt{6}$

is increasing at every point of its domain is 

$f^{\mathrm{\prime }}\left(x\right)=3\left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^2+5>0$ If

$\begin{array}{r}{x}^2>\frac{-5}{3\left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)}\text{ if }\\ 1-\frac{\sqrt{21-4b-b^2}}{b+1}>0\end{array}$

which is trivially true if $b+1<0\text{ i.e. }b<-1$

If $b+1>0$ then

$(b+1{)}^2>21-4b-b^2\Rightarrow (b\mid -2\left)\right(b+5)>0$

$\text{ i.e. }b>2\text{ or }b<-5$

so $b\in [2,3]$