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

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(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. We differentiate $f(x) = 	an^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}ight)$ with respect t

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$2$

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(C) Let $I = \int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx$. We differentiate $f(x) = 	an^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}ight)$ with respect to $x$. Using the chain rule,$\frac{d}{dx}\left(	an^{-1}(u)ight) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$,where $u = \sqrt{\frac{x^2+x+1}{x}}$. Then $u^2 = \frac{x^2+x+1}{x} = x + 1 + \frac{1}{x}$. $\frac{du}{dx} = \frac{1}{2\sqrt{u^2}} \cdot \frac{d}{dx}\left(x + 1 + \frac{1}{x}ight) = \frac{1}{2u} \left(1 - \frac{1}{x^2}ight) = \frac{1}{2u} \left(\frac{x^2-1}{x^2}ight)$. So,$\frac{d}{dx}\left(	an^{-1}(u)ight) = \frac{1}{1 + \frac{x^2+x+1}{x}} \cdot \frac{x^2-1}{2x^2 \sqrt{\frac{x^2+x+1}{x}}} = \frac{x}{x^2+2x+1} \cdot \frac{x^2-1}{2x^2 \frac{\sqrt{x^2+x+1}}{\sqrt{x}}} = \frac{x^2-1}{(x+1)^2 \cdot 2x \sqrt{x(x^2+x+1)}} = \frac{1}{2} \cdot \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}}$. Thus,$\int \frac{x^2-1}{(x+1)^2 \sqrt{x(x^2+x+1)}} dx = 2 	an^{-1}\left(\sqrt{\frac{x^2+x+1}{x}}ight) + C$. Comparing this with the given expression,we get $A = 2$.

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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