find $b$ if function $f(x)$ is strictly decreasing function if $\displaystyle f(x) = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$ is strictly decreasing
function forall real values of $x.$ then $b$
$\displaystyle f'(x) = 2\sqrt{3} \; \frac{1}{3+(2x+1)^2}-\frac{2x+1}{x^2+x+1}+(b^2-5b+3)$
function $f(x)$ is strictly decreasing if $f'(x)<0$ for all real $x$
$\displaystyle 2\sqrt{3} \; \frac{1}{3+(2x+1)^2}-\frac{2x+1}{x^2+x+1}+(b^2-5b+3)<0$
could some help me with this, thanks

I think that you made a mistake in the derivative $$f(x)=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$$ should lead, after basic simplifications,  to $$f'(x)=(b^2-5b+3)-\frac{2 x}{x^2+x+1}$$ Since we are in the real domain, $(x^2+x+1)$ does not show any real root, then $f'x)<0$ reduces to $$(b^2-5b+3)(x^2+x+1)-2x <0$$
I am sure that you can take it from here.
