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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $\frac{x^2+x+1}{x^2+2x+1} = A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$ Multiply both sides by $(x+1)^2$: $x^2+x+1 = A(x+1)^2 + B(x

Vedclass · Accepted Answer

$2C$

Vedclass · Answer

(D) Given the equation: $\frac{x^2+x+1}{x^2+2x+1} = A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$ Multiply both sides by $(x+1)^2$: $x^2+x+1 = A(x+1)^2 + B(x+1) + C$ $x^2+x+1 = A(x^2+2x+1) + Bx + B + C$ $x^2+x+1 = Ax^2 + (2A+B)x + (A+B+C)$ Comparing the coefficients of $x^2$,$x$,and the constant term: $A = 1$ $2A+B = 1$ $\Rightarrow 2(1)+B = 1$ $\Rightarrow B = -1$ $A+B+C = 1$ $\Rightarrow 1-1+C = 1$ $\Rightarrow C = 1$ Now,calculate $A-B$: $A-B = 1 - (-1) = 2$ Since $C = 1$,we have $2 = 2C$. Therefore,$A-B = 2C$.

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

Similar Questions

$\frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \Rightarrow 7A + 2B + C = ?$

$\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = $

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2A-3B+C=$

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

Which of the following is a partial fraction of $\frac{-x^{2} + 6x + 13}{(3x + 5)(x^{2} + 4x + 4)} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module