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

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(C) Given that, $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{(2x-1)}+\frac{B}{(3x+1)}$Multiplying both sides by $(2x-1)(3x+1)$, we get:$x+1 = A(3x+1) + B(2x-1)$

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$6$

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(C) Given that, $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{(2x-1)}+\frac{B}{(3x+1)}$Multiplying both sides by $(2x-1)(3x+1)$, we get:$x+1 = A(3x+1) + B(2x-1)$$x+1 = x(3A+2B) + (A-B)$Comparing the coefficients of $x$ and the constant terms on both sides, we get:$3A+2B = 1$  ...$(i)$$A-B = 1$  ...$(ii)$From equation $(ii)$, $A = B+1$. Substituting this into equation $(i)$:$3(B+1) + 2B = 1$$3B+3+2B = 1$$5B = -2 \Rightarrow B = -\frac{2}{5}$Now, $A = -\frac{2}{5} + 1 = \frac{3}{5}$We need to find $16A+9B$:$16\left(\frac{3}{5}\right) + 9\left(-\frac{2}{5}\right) = \frac{48}{5} - \frac{18}{5} = \frac{30}{5} = 6$

If $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{(2x-1)}+\frac{B}{(3x+1)}$, then $16A+9B$ is equal to

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