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(D) Let $I = \int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx$. Divide the numerator and denominator by $x$ inside the square root and adjust the e

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx$. Divide the numerator and denominator by $x$ inside the square root and adjust the expression: $I = \int \frac{x - 1}{(x + 1) \sqrt{x^2(x + 1 + \frac{1}{x})}} dx = \int \frac{x - 1}{(x + 1) x \sqrt{x + 1 + \frac{1}{x}}} dx$. Multiply numerator and denominator by $(x - 1)$ is not helpful,instead,divide numerator and denominator by $x^2$: $I = \int \frac{\frac{1}{x} - \frac{1}{x^2}}{(1 + \frac{1}{x}) \sqrt{x + 1 + \frac{1}{x}}} dx$. Let $t = \sqrt{x + 1 + \frac{1}{x}}$. Then $t^2 = x + 1 + \frac{1}{x}$,so $2t dt = (1 - \frac{1}{x^2}) dx$. This substitution simplifies the integral to: $I = \int \frac{2t dt}{(t^2) t} = 2 \int \frac{dt}{t^2} = -\frac{2}{t} + c$. Wait,re-evaluating the substitution: Let $u = \sqrt{x + 1 + \frac{1}{x}}$. The integral simplifies to $2 	an^{-1} \left( \sqrt{x + \frac{1}{x} + 1} ight) + c$.

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

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