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

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(C) Given, $\int \frac{x+3}{(x-1)^2(2 x-1)} d x=\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+K$. Using partial fractions, we write: $\frac{x+3}{(x-1)^2(2

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$-4$

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(C) Given, $\int \frac{x+3}{(x-1)^2(2 x-1)} d x=\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+K$. Using partial fractions, we write: $\frac{x+3}{(x-1)^2(2 x-1)}=\frac{\alpha}{x-1}+\frac{\beta}{(x-1)^2}+\frac{\gamma}{2 x-1}$. Multiplying by $(x-1)^2(2 x-1)$, we get: $x+3=\alpha(x-1)(2 x-1)+\beta(2 x-1)+\gamma(x-1)^2$. Setting $x=1$, we get $4=\beta(2-1) \Rightarrow \beta=4$. Setting $x=1/2$, we get $3.5=\gamma(1/2-1)^2 \Rightarrow 3.5=\gamma(1/4) \Rightarrow \gamma=14$. Comparing the coefficient of $x^2$, we get $0=2\alpha+\gamma \Rightarrow 2\alpha=-14 \Rightarrow \alpha=-7$. Integrating: $\int \left( \frac{-7}{x-1}+\frac{4}{(x-1)^2}+\frac{14}{2 x-1} \right) d x = -7 \log |x-1| - \frac{4}{x-1} + \frac{14}{2} \log |2 x-1| + K = -7 \log |x-1| - \frac{4}{x-1} + 7 \log |2 x-1| + K$. Comparing with the given form, we have $A=-4$, $B=7$, and $C=-7$. Therefore, $A+B+C = -4+7-7 = -4$.

If $\int \frac{x+3}{(x-1)^2(2 x-1)} d x=\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+K$, then $A+B+C=$

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