$\begin{aligned} & \text { Let } I=\int \frac{x^2-1}{(x+1)^2 \sqrt{x\left(x^2+x+1\right)}} d x \\ & \text { Now, } \frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}\right) \\ & =\frac{1}{1+\frac{x^2+x+1}{x}} \times \frac{d}{d x}\left(\sqrt{\frac{x^2+x+1}{x}}\right) \\ & =\frac{x}{x^2+2 x+1}\left[\frac{\sqrt{x} \times \frac{2 x+1}{2 \sqrt{x^2+x+1}}-\sqrt{x^2+x+1} \frac{1}{2 \sqrt{x}}}{(\sqrt{x})^2}\right] \\ & =\frac{x}{x^2+2 x+1}\left[\frac{\sqrt{x}(2 x+1)}{2 \sqrt{\left(x^2+x+1\right)}}-\frac{\sqrt{x^2+x+1}}{2 \sqrt{x}}\right] \\ & =\frac{x}{\left(x^2+2 x+1\right)}\left[\frac{(2 x+1) x-\left(x^2+x+1\right)}{\frac{2 \sqrt{x\left(x^2+x+1\right)}}{x}}\right] \\ & =\frac{x}{(x+1)^2}\left[\frac{x^2-1}{2 x \sqrt{x\left(x^2+x+1\right)}}\right] \\ & =\frac{x^2-1}{2(x+1)^2 \sqrt{x\left(x^2+x+1\right)}} \\ & \end{aligned}$
$\begin{aligned} \therefore I & =\int \frac{d}{d x}\left(2 \tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}\right) \\ & =2 \tan ^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}\right)+C \\ \therefore A & =2\end{aligned}$