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

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

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$2$

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(B) Given the partial fraction decomposition: $\frac{3 x+2}{(x+1)(2 x^2+3)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3}$ Multiplying both sides by $(x+1)(2 x^2+3)$,we get: $3 x+2=A(2 x^2+3)+(B x+C)(x+1)$ To find $A$,put $x=-1$: $3(-1)+2=A(2(-1)^2+3) \Rightarrow -1=A(5) \Rightarrow A=-\frac{1}{5}$ Expanding the right side: $3 x+2=2 A x^2+3 A+B x^2+B x+C x+C$ $3 x+2=(2 A+B) x^2+(B+C) x+(3 A+C)$ Comparing the coefficients of $x^2$: $2 A+B=0 \Rightarrow B=-2 A=-2(-\frac{1}{5})=\frac{2}{5}$ Comparing the coefficients of $x$: $B+C=3 \Rightarrow C=3-B=3-\frac{2}{5}=\frac{13}{5}$ Finally,calculate $A+C-B$: $-\frac{1}{5}+\frac{13}{5}-\frac{2}{5}=\frac{13-2-1}{5}=\frac{10}{5}=2$

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

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