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 $\frac{x}{(x+1)^2}$ as 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 $\frac{x}{(x+1)^2}$ as partial fractions:$\frac{x}{(x+1)^2} = \frac{P}{x+1} + \frac{Q}{(x+1)^2}$.$x = P(x+1) + Q = Px + (P+Q)$.Comparing coefficients,$P=1$ and $P+Q=0$,so $Q=-1$.Thus,$\frac{x^2+x+1}{x^2+2x+1} = 1 - (\frac{1}{x+1} - \frac{1}{(x+1)^2}) = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}$.Comparing 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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