int frac{x , dx}{(x-1)(x-2)} equals | vedclass

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(D) Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.By partial fraction decomposition,$x = A(x-2) + B(x-1)$.Setting $x = 1$,we get $1 = A(1

Vedclass · Accepted Answer

$\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$

Vedclass · Answer

(D) Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.By partial fraction decomposition,$x = A(x-2) + B(x-1)$.Setting $x = 1$,we get $1 = A(1-2) \Rightarrow A = -1$.Setting $x = 2$,we get $2 = B(2-1) \Rightarrow B = 2$.Thus,$\frac{x}{(x-1)(x-2)} = -\frac{1}{x-1} + \frac{2}{x-2}$.Integrating both sides with respect to $x$:$\int \frac{x \, dx}{(x-1)(x-2)} = \int \left( -\frac{1}{x-1} + \frac{2}{x-2} \right) dx$.$= -\log |x-1| + 2 \log |x-2| + C$.$= \log |x-2|^2 - \log |x-1| + C$.$= \log \left| \frac{(x-2)^2}{x-1} \right| + C$.Therefore,the correct option is $D$.

$\int \frac{x \, dx}{(x-1)(x-2)}$ equals

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