If x neq -1 and int frac{x^3+x^2-x-1}{(x^5+x^ | vedclass

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Question

(A) Let $I = \int \frac{x^3+x^2-x-1}{(x^5+x^4+3x^3+3x^2+x+1) 	an^{-1}(\frac{x^2+1}{x})} dx$. Factor the numerator: $x^2(x+1) - 1(x+1) = (x^2-1)(x+1)

Vedclass · Accepted Answer

$\frac{-3}{2}$

Vedclass · Answer

(A) Let $I = \int \frac{x^3+x^2-x-1}{(x^5+x^4+3x^3+3x^2+x+1) 	an^{-1}(\frac{x^2+1}{x})} dx$. Factor the numerator: $x^2(x+1) - 1(x+1) = (x^2-1)(x+1) = (x-1)(x+1)^2$. Factor the denominator: $x^4(x+1) + 3x^2(x+1) + 1(x+1) = (x+1)(x^4+3x^2+1)$. So, $I = \int \frac{(x-1)(x+1)^2}{(x+1)(x^4+3x^2+1) 	an^{-1}(\frac{x^2+1}{x})} dx = \int \frac{(x^2-1)}{(x^4+3x^2+1) 	an^{-1}(\frac{x^2+1}{x})} dx$. Divide numerator and denominator by $x^2$: $I = \int \frac{1 - \frac{1}{x^2}}{(x^2 + 3 + \frac{1}{x^2}) 	an^{-1}(x + \frac{1}{x})} dx = \int \frac{1 - \frac{1}{x^2}}{((x + \frac{1}{x})^2 + 1) 	an^{-1}(x + \frac{1}{x})} dx$. Let $u = 	an^{-1}(x + \frac{1}{x})$, then $du = \frac{1}{1 + (x + \frac{1}{x})^2} \cdot (1 - \frac{1}{x^2}) dx$. Thus, $I = \int \frac{1}{u} du = \log|u| + C = \log|	an^{-1}(x + \frac{1}{x})| + C$. Comparing with $A \log(f(x)) + C$, we get $A = 1$ and $f(x) = 	an^{-1}(x + \frac{1}{x})$. Then $f(2) = 	an^{-1}(2 + \frac{1}{2}) = 	an^{-1}(\frac{5}{2})$. Therefore, $A - 	an(f(2)) = 1 - 	an(	an^{-1}(\frac{5}{2})) = 1 - \frac{5}{2} = \frac{-3}{2}$.

If $x \neq -1$ and $\int \frac{x^3+x^2-x-1}{(x^5+x^4+3x^3+3x^2+x+1) \tan^{-1}(\frac{x^2+1}{x})} dx = A \log(f(x)) + C$, then $A - \tan(f(2)) = $

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