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

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(D) Let $I = \int \frac{x-1}{(x+1) \sqrt{x(x^2+x+1)}} dx$.Divide numerator and denominator by $x$:$I = \int \frac{1 - \frac{1}{x}}{(x+1) \sqrt{x^2+x+1

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$2 \cdot 	an^{-1}\left(\sqrt{x+\frac{1}{x}+1}ight)+c$

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(D) Let $I = \int \frac{x-1}{(x+1) \sqrt{x(x^2+x+1)}} dx$.Divide numerator and denominator by $x$:$I = \int \frac{1 - \frac{1}{x}}{(x+1) \sqrt{x^2+x+1}} dx$.Wait,let us simplify the expression inside the square root by dividing by $x^2$:$I = \int \frac{x-1}{(x+1) \sqrt{x^2(1 + \frac{1}{x} + \frac{1}{x^2})}} dx = \int \frac{x-1}{x(x+1) \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}} dx$.This can be rewritten as:$I = \int \frac{1 - \frac{1}{x}}{(x+1) \sqrt{x^2+x+1}} dx$.Actually,the standard substitution method for this integral is:Let $t = \sqrt{x + \frac{1}{x} + 1}$.Then $t^2 = x + \frac{1}{x} + 1$.Differentiating both sides: $2t dt = (1 - \frac{1}{x^2}) dx$.The integral becomes $\int \frac{1 - \frac{1}{x^2}}{x + \frac{1}{x} + 2} \cdot \frac{1}{\sqrt{x + \frac{1}{x} + 1}} dx$.Substituting $t$: $\int \frac{2 dt}{t^2 + 1} = 2 	an^{-1}(t) + c$.Thus,$I = 2 	an^{-1}\left(\sqrt{x + \frac{1}{x} + 1}ight) + c$.

$\int \frac{x-1}{(x+1) \sqrt{x(x^2+x+1)}} dx = $

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