If int frac{(x^2-1)}{(x+1)^2 sqrt{x(x^2+x+1)} | vedclass

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(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. Consider the derivative of the right-hand side: $\frac{d}{dx} \left( A 	an^{-1} \sqrt{

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$2$

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(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. Consider the derivative of the right-hand side: $\frac{d}{dx} \left( A 	an^{-1} \sqrt{\frac{x^2+x+1}{x}} ight) = A \cdot \frac{1}{1 + \frac{x^2+x+1}{x}} \cdot \frac{d}{dx} \left( \sqrt{\frac{x^2+x+1}{x}} ight)$. $= A \cdot \frac{x}{x^2+2x+1} \cdot \frac{1}{2 \sqrt{\frac{x^2+x+1}{x}}} \cdot \left( \frac{x(2x+1) - (x^2+x+1)}{x^2} ight)$. $= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \left( \frac{2x^2+x-x^2-x-1}{x^2} ight)$. $= A \cdot \frac{x}{(x+1)^2} \cdot \frac{\sqrt{x}}{2 \sqrt{x^2+x+1}} \cdot \frac{x^2-1}{x^2}$. $= A \cdot \frac{x^2-1}{2(x+1)^2 \sqrt{x(x^2+x+1)}}$. Comparing this with the integrand, we have $\frac{A}{2} = 1$, which implies $A = 2$.

If $\int \frac{(x^2-1)}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx = A \tan^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}\right) + C$, where $C$ is a constant, then $A$ equals to

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