int {frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} | vedclass

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Question

(B) We have the integral $I = \int {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} \,dx$. First,perform polynomial division: $\frac{{{x^2} + x - 1}}{{{x^2}

Vedclass · Accepted Answer

$x - \log |x + 3| + \log |x - 2| + c$

Vedclass · Answer

(B) We have the integral $I = \int {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} \,dx$. First,perform polynomial division: $\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}} = \frac{({x^2} + x - 6) + 5}{{{x^2} + x - 6}} = 1 + \frac{5}{{{x^2} + x - 6}}$. Now,factor the denominator: ${x^2} + x - 6 = (x + 3)(x - 2)$. Using partial fractions: $\frac{5}{{(x + 3)(x - 2)}} = \frac{A}{x + 3} + \frac{B}{x - 2}$. $5 = A(x - 2) + B(x + 3)$. Setting $x = 2$,we get $5 = 5B \implies B = 1$. Setting $x = -3$,we get $5 = -5A \implies A = -1$. Thus,$\frac{5}{{(x + 3)(x - 2)}} = \frac{1}{x - 2} - \frac{1}{x + 3}$. Integrating: $I = \int {\left( {1 + \frac{1}{x - 2} - \frac{1}{x + 3}} \right)} \,dx = x + \log |x - 2| - \log |x + 3| + c$.

$\int {\frac{{{x^2} + x - 1}}{{{x^2} + x - 6}}} \,dx = $

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