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

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(A) Given integral: $I = \int \frac{x^{2}+1}{(x-3)(x-2)} d x$Since the degree of the numerator is equal to the degree of the denominator,we perform di

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$1, 10, -5$

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(A) Given integral: $I = \int \frac{x^{2}+1}{(x-3)(x-2)} d x$Since the degree of the numerator is equal to the degree of the denominator,we perform division:$\frac{x^{2}+1}{x^{2}-5x+6} = 1 + \frac{5x-5}{x^{2}-5x+6}$So,$I = \int (1 + \frac{5x-5}{(x-3)(x-2)}) d x = \int 1 d x + 5 \int \frac{x-1}{(x-3)(x-2)} d x$Using partial fractions for $\frac{x-1}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$:$x-1 = A(x-2) + B(x-3)$For $x=3$: $3-1 = A(3-2) \Rightarrow A=2$For $x=2$: $2-1 = B(2-3) \Rightarrow B=-1$Thus,$I = x + 5 \int (\frac{2}{x-3} - \frac{1}{x-2}) d x = x + 10 \log |x-3| - 5 \log |x-2| + c$Comparing with $Px + Q \log |x-3| + R \log |x-2| + c$,we get $P=1, Q=10, R=-5$.

$\int \frac{x^{2}+1}{(x-3)(x-2)} d x = P x + Q \log |x-3| + R \log |x-2| + c$,where $c$ is the constant of integration. Then the values of $P, Q, R$ are,respectively:

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