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

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(A) Let $I = \int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x$. Divide numerator and denominator by $x^2$ inside the square root: $I = \int \frac{x-1}{(x+1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x$. Divide numerator and denominator by $x^2$ inside the square root: $I = \int \frac{x-1}{(x+1) \sqrt{x^2(x+1+\frac{1}{x})}} d x = \int \frac{x-1}{x(x+1) \sqrt{x+1+\frac{1}{x}}} d x$. Multiply numerator and denominator by $x$: $I = \int \frac{x^2-1}{x^2(x+1) \sqrt{x+1+\frac{1}{x}}} d x$. This can be rewritten as: $I = \int \frac{(1 - \frac{1}{x^2}) d x}{(1 + \frac{1}{x}) \sqrt{x+1+\frac{1}{x}}}$. Let $t = \sqrt{x+1+\frac{1}{x}}$. Then $t^2 = x+1+\frac{1}{x}$,so $2t dt = (1 - \frac{1}{x^2}) dx$. Also,$t^2 - 1 = x + \frac{1}{x} = \frac{x^2+1}{x}$. This substitution is slightly complex,so let's use $t = \sqrt{x+1+\frac{1}{x}}$ directly. $I = \int \frac{2t dt}{(1 + \frac{1}{x}) t} = 2 \int \frac{dt}{1 + \frac{1}{x}}$. Actually,the standard substitution for this form is $t = \sqrt{\frac{x^2+x+1}{x}}$. Then $t^2 = x+1+\frac{1}{x}$,$2t dt = (1-\frac{1}{x^2}) dx$. $I = \int \frac{(1-1/x^2) dx}{(1+1/x) \sqrt{x+1+1/x}} = \int \frac{2t dt}{(1+1/x)t} = 2 \int \frac{dt}{1+1/x}$. Following the provided steps: $I = 2 	an^{-1} \left( \sqrt{\frac{x^2+x+1}{x}} ight) + c$.

$\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}} d x=$

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