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

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

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$6$

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(C) Given the partial fraction decomposition: $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{2x-1}+\frac{B}{3x+1}$ Multiplying both sides by $(2x-1)(3x+1)$,we get: $x+1=A(3x+1)+B(2x-1)$ To find $A$,substitute $x=\frac{1}{2}$: $\frac{1}{2}+1=A(3(\frac{1}{2})+1)+B(0)$ $\Rightarrow \frac{3}{2}=A(\frac{5}{2})$ $\Rightarrow A=\frac{3}{5}$ To find $B$,substitute $x=-\frac{1}{3}$: $-\frac{1}{3}+1=A(0)+B(2(-\frac{1}{3})-1)$ $\Rightarrow \frac{2}{3}=B(-\frac{5}{3})$ $\Rightarrow B=-\frac{2}{5}$ Now,calculate $16A+9B$: $16(\frac{3}{5})+9(-\frac{2}{5}) = \frac{48}{5}-\frac{18}{5} = \frac{30}{5} = 6$ Thus,the value is $6$.

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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