If \(\frac{x^2+x+1}{x^2+2 x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}\), then \(A-B\) is equal to
If \(\frac{x^2+x+1}{x^2+2 x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}\), then \(A-B\) is equal to
\(4 C\)
\(4 C+1\)
\(3 C\)
\(2 \mathrm{C}\)
Solution
It is given that,
\(\begin{aligned}
& \frac{x^2+x+1}{x^2+2 x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2} \\
\Rightarrow \quad & x^2+x+1=A\left(x^2+2 x+1\right)+B(x+1)+C \\
\Rightarrow \quad & x^2+x+1=A x^2+(2 A+B) x+(A+B+C)
\end{aligned}\)
On comparing the coefficient of different terms, we get \(A=1,2 A+B=1\) and \(A+B+C=1\)
\(\begin{aligned}
\therefore & B & =-1, \text { and } C=1 \\
\therefore & A-B & =2=2 C.
\end{aligned}\)