If frac{x^2+x+1}{x^2+2x+1} = A + frac{B}{x+1} | vedclass

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(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$. Now,express $\frac{x}{(x+1)^2}$ in partial fractions

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$2C$

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(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$. Now,express $\frac{x}{(x+1)^2}$ in partial fractions: $\frac{x}{(x+1)^2} = \frac{A'}{x+1} + \frac{B'}{(x+1)^2} = \frac{A'(x+1) + B'}{(x+1)^2}$. Comparing numerators,$x = A'x + (A'+B')$. Thus,$A' = 1$ and $A'+B' = 0$,which gives $B' = -1$. So,$\frac{x^2+x+1}{x^2+2x+1} = 1 - (\frac{1}{x+1} - \frac{1}{(x+1)^2}) = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}$. Comparing this with $A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$,we get $A=1$,$B=-1$,and $C=1$. Therefore,$A-B = 1 - (-1) = 2$. Since $C=1$,$2 = 2C$. Hence,$A-B = 2C$.

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

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