If int frac{3x+1}{(x-1)(x-2)(x-3)} dx = A log | vedclass

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(D) We have the integral $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx$. Using partial fractions,let $\frac{3x+1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{

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$2, -7, 5$

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(D) We have the integral $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx$. Using partial fractions,let $\frac{3x+1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$. Multiplying both sides by $(x-1)(x-2)(x-3)$,we get: $3x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$. Setting $x=1$: $3(1)+1 = A(1-2)(1-3) \Rightarrow 4 = A(-1)(-2) \Rightarrow 4 = 2A \Rightarrow A = 2$. Setting $x=2$: $3(2)+1 = B(2-1)(2-3) \Rightarrow 7 = B(1)(-1) \Rightarrow 7 = -B \Rightarrow B = -7$. Setting $x=3$: $3(3)+1 = C(3-1)(3-2) \Rightarrow 10 = C(2)(1) \Rightarrow 10 = 2C \Rightarrow C = 5$. Thus,the integral becomes $\int (\frac{2}{x-1} - \frac{7}{x-2} + \frac{5}{x-3}) dx$. Integrating term by term,we get $2 \log |x-1| - 7 \log |x-2| + 5 \log |x-3| + K$. Comparing this with $A \log |x-1| + B \log |x-2| + C \log |x-3| + K$,we find $A=2, B=-7, C=5$.

If $\int \frac{3x+1}{(x-1)(x-2)(x-3)} dx = A \log |x-1| + B \log |x-2| + C \log |x-3| + K$,then the values of $A, B$,and $C$ are respectively:

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