If $\int \frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}=A{\mathrm{Tan}}^{-1}\sqrt{f\left(x\right)}+$ constant, then the ordered pair $(A,f(-1\left)\right)=$

A.${\mathrm{Tan}}^{-1}\sqrt{f\left(x\right)}+c=\int \frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}=\int \frac{\left(x^2-1\right)dx}{(x+1{)}^2\sqrt{x^2(x+1+1/x)}}$

$\begin{array}{l}=\int \frac{\left(x^2-1\right)dx}{\left(x^2+2x+1\right)x\sqrt{x+\frac{1}{x}+1}}=\int \frac{x^2\left(1-\frac{1}{x^2}\right)dx}{x^2\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}=2{\mathrm{Tan}}^{-1}\sqrt{x+\frac{1}{x}+1}+c\\ =\int \frac{2tdt}{\left(t^2+1\right)t}\text{ where }{t}^2=x+\frac{1}{x}+1\text{ and }2tdt=\left(x-\frac{1}{x^2}\right)dx\\ =2{\mathrm{Tan}}^{-1}\sqrt{x+\frac{1}{x}+1}+c\\ \Rightarrow A=2,f\left(x\right)=x+\frac{1}{x}+1.\text{ Now }f(-1)=-1-1+1=-1.\therefore (A,f(-1\left)\right)=(2,-1).\end{array}$