If f(x) = frac{3-8x}{3x-1} and int f(y) dy = | vedclass

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(D) Given $f(y) = \frac{3-8y}{3y-1}$.We need to evaluate $\int \frac{3-8y}{3y-1} dy$.Perform polynomial division or algebraic manipulation:$\frac{3-8y

Vedclass · Accepted Answer

$\frac{-3}{2}$

Vedclass · Answer

(D) Given $f(y) = \frac{3-8y}{3y-1}$.We need to evaluate $\int \frac{3-8y}{3y-1} dy$.Perform polynomial division or algebraic manipulation:$\frac{3-8y}{3y-1} = \frac{-\frac{8}{3}(3y-1) + \frac{8}{3} + 3}{3y-1} = \frac{-\frac{8}{3}(3y-1) + \frac{17}{3}}{3y-1} = -\frac{8}{3} + \frac{17}{3(3y-1)}$.Now, integrate with respect to $y$:$\int (-\frac{8}{3} + \frac{17}{3(3y-1)}) dy = -\frac{8}{3}y + \frac{17}{3} \cdot \frac{1}{3} \log |3y-1| + C = -\frac{8}{3}y + \frac{17}{9} \log |3y-1| + C$.Comparing this with $Ay + B \log |3y-1| + C$, we get $A = -\frac{8}{3}$ and $B = \frac{17}{9}$.Then, $\frac{A-3B}{2} = \frac{-\frac{8}{3} - 3(\frac{17}{9})}{2} = \frac{-\frac{8}{3} - \frac{17}{3}}{2} = \frac{-\frac{25}{3}}{2} = -\frac{25}{6}$.Wait, re-evaluating the original expression: $\frac{3-8y}{3y-1} = \frac{-\frac{8}{3}(3y-1) + 3 - \frac{8}{3}}{3y-1} = -\frac{8}{3} + \frac{1/3}{3y-1}$.Integrating: $-\frac{8}{3}y + \frac{1}{9} \log |3y-1| + C$.So $A = -\frac{8}{3}$ and $B = \frac{1}{9}$.$\frac{A-3B}{2} = \frac{-\frac{8}{3} - 3(\frac{1}{9})}{2} = \frac{-\frac{8}{3} - \frac{1}{3}}{2} = \frac{-3}{2}$.

If $f(x) = \frac{3-8x}{3x-1}$ and $\int f(y) dy = Ay + B \log |3y-1| + C$, then $\frac{A-3B}{2} =$

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