If fleft( frac{x - 4}{x + 2} right) = 2x + 1 | vedclass

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(B) Let $y = \frac{x - 4}{x + 2}$.Then $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.Substit

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$-12 \log_e |1 - x| - 3x + C$

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(B) Let $y = \frac{x - 4}{x + 2}$.Then $x - 4 = y(x + 2) \Rightarrow x - 4 = xy + 2y$.$x(1 - y) = 2y + 4 \Rightarrow x = \frac{2y + 4}{1 - y}$.Substituting $x$ into $f(y) = 2x + 1$,we get $f(y) = 2\left( \frac{2y + 4}{1 - y} \right) + 1$.$f(y) = \frac{4y + 8 + 1 - y}{1 - y} = \frac{3y + 9}{1 - y}$.Thus,$f(x) = \frac{3x + 9}{1 - x} = \frac{3(x - 1 + 4)}{1 - x} = \frac{3(x - 1)}{1 - x} + \frac{12}{1 - x} = -3 + \frac{12}{1 - x}$.Now,$\int f(x) \,dx = \int \left( -3 + \frac{12}{1 - x} \right) \,dx$.$= -3x + 12 \int \frac{1}{1 - x} \,dx$.$= -3x + 12 \left( \frac{\log_e |1 - x|}{-1} \right) + C$.$= -12 \log_e |1 - x| - 3x + C$.

If $f\left( \frac{x - 4}{x + 2} \right) = 2x + 1$ for $x \in R \setminus \{ -2 \}$,then $\int f(x) \,dx$ is equal to (where $C$ is a constant of integration)

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