If int frac{(x-1) dx}{(x+1) sqrt{x^3+x^2+x}} | vedclass

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(B) We have, $I = \int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}}$.Dividing numerator and denominator by $x$, we get $I = \int \frac{(1 - 1/x) dx}{(1 + 1

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$(2, -1)$

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(B) We have, $I = \int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}}$.Dividing numerator and denominator by $x$, we get $I = \int \frac{(1 - 1/x) dx}{(1 + 1/x) \sqrt{x + 1 + 1/x}}$.Let $t = \sqrt{x + 1 + 1/x}$. Then $t^2 = x + 1 + 1/x$.Differentiating both sides, $2t dt = (1 - 1/x^2) dx = \frac{x^2-1}{x^2} dx = \frac{(x-1)(x+1)}{x^2} dx$.Also, $1 + 1/x = \frac{x+1}{x}$.Substituting these into the integral, we get $I = \int \frac{2t dt}{t^2 \cdot (t^2-1) \cdot \frac{x}{x+1} \cdot \frac{x+1}{x}} = \int \frac{2 dt}{t^2+1} = 2 	an^{-1}(t) + C$.Thus, $I = 2 	an^{-1} \sqrt{x + 1 + 1/x} + C$.Comparing with $A 	an^{-1} \sqrt{f(x)} + C$, we get $A = 2$ and $f(x) = x + 1 + 1/x$.Then $f(-1) = -1 + 1 + (1/-1) = -1$.Therefore, the ordered pair $(A, f(-1)) = (2, -1)$.

If $\int \frac{(x-1) dx}{(x+1) \sqrt{x^3+x^2+x}} = A \cdot \tan^{-1} \sqrt{f(x)} + \text{constant}$, then the ordered pair $(A, f(-1)) =$

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