If $f(x)=a{\tan }^{-1}x+2b\log (1+x)+x+1$ has critical points at $x=0$ and $x=2$, then the values of the constants $a$ and $b$ are respectively:

Find $f^{\prime }(x)$:
$$f^{\prime }(x)=\frac{a}{1+x^2}+\frac{2b}{1+x}+1$$
Critical points occur where $f^{\prime }(x)=0$.At $x=0$:
$$f^{\prime }(0)=\frac{a}{1}+\frac{2b}{1}+1=0⟹a+2b=-1$$
--- (1)At $x=2$:
$$f^{\prime }(2)=\frac{a}{1+4}+\frac{2b}{1+2}+1=0$$
$$\frac{a}{5}+\frac{2b}{3}+1=0⟹3a+10b=-15$$
--- (2)Multiply (1) by 3: $3a+6b=-3$.Subtract from (2): $(3a+10b)-(3a+6b)=-15-(-3)$ $
$4b=-12⟹b=-3$$Substitute$ b$into(1):$ a + 2(-3) = -1 \implies a - 6 = -1 \implies a = 5$.So$ a = 5$and$ b = -3$.