If int {frac{{2x + 3}}{{{x^2} - 5x + 6}}} ;dx | vedclass

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(C) We are given the integral: $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx$. First,we express the numerator in terms of the derivative of the denom

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$	ext{Constant}$

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(C) We are given the integral: $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx$. First,we express the numerator in terms of the derivative of the denominator: $\frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$. So,$\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5 + 8}}{{{x^2} - 5x + 6}}} \;dx = \int {\frac{{2x - 5}}{{{x^2} - 5x + 6}}} \;dx + \int {\frac{8}{{(x - 2)(x - 3)}}} \;dx$. Using partial fractions for the second part: $\frac{8}{(x - 2)(x - 3)} = \frac{8}{x - 3} - \frac{8}{x - 2}$. Thus,the integral becomes: $\ln |x^2 - 5x + 6| + 8 \int (\frac{1}{x - 3} - \frac{1}{x - 2}) \;dx + C$. $= \ln |(x - 2)(x - 3)| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$. $= \ln |x - 2| + \ln |x - 3| + 8 \ln |x - 3| - 8 \ln |x - 2| + C$. $= 9 \ln |x - 3| - 7 \ln |x - 2| + C$. Comparing this with the given expression $9 \ln (x - 3) - 7 \ln (x - 2) + A$,we find that $A$ must be an arbitrary constant $C$.

If $\int {\frac{{2x + 3}}{{{x^2} - 5x + 6}}} \;dx = 9\ln (x - 3) - 7\ln (x - 2) + A$,then $A = $

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