If frac{x}{(1+x^2)(3-2x)} = frac{Bx+C}{1+x^2} | vedclass

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(D) Given the partial fraction decomposition: $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$.Multiplying both sides by $(1+x^2)(3-2x)

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$\frac{-2}{13}$

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(D) Given the partial fraction decomposition: $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$.Multiplying both sides by $(1+x^2)(3-2x)$,we get:$x = (Bx+C)(3-2x) + A(1+x^2)$$x = 3Bx - 2Bx^2 + 3C - 2Cx + A + Ax^2$$x = (A-2B)x^2 + (3B-2C)x + (A+3C)$Comparing the coefficients on both sides:$1$) $A - 2B = 0 \Rightarrow A = 2B$$2$) $3B - 2C = 1$$3$) $A + 3C = 0 \Rightarrow A = -3C$From $A = 2B$ and $A = -3C$,we have $2B = -3C \Rightarrow B = -\frac{3}{2}C$.Substitute $B$ into equation $(2)$:$3(-\frac{3}{2}C) - 2C = 1$$-\frac{9}{2}C - 2C = 1$$-\frac{13}{2}C = 1$$C = -\frac{2}{13}$.

If $\frac{x}{(1+x^2)(3-2x)} = \frac{Bx+C}{1+x^2} + \frac{A}{3-2x}$,then $C$ is

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