If the partial fraction decomposition of frac | vedclass

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

Question

(B) Given the expression $\frac{x^2+1}{x^3+3x^2+3x+2} = \frac{x^2+1}{(x+2)(x^2+x+1)}$.We write the partial fraction decomposition as $\frac{x^2+1}{(x+

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the expression $\frac{x^2+1}{x^3+3x^2+3x+2} = \frac{x^2+1}{(x+2)(x^2+x+1)}$.We write the partial fraction decomposition as $\frac{x^2+1}{(x+2)(x^2+x+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$.Multiplying both sides by $(x+2)(x^2+x+1)$,we get $x^2+1 = A(x^2+x+1) + (Bx+C)(x+2)$.Setting $x = -2$: $(-2)^2 + 1 = A(4-2+1)$ $\Rightarrow 5 = 3A$ $\Rightarrow A = \frac{5}{3}$.Comparing the coefficients of $x^2$: $1 = A + B \Rightarrow B = 1 - \frac{5}{3} = -\frac{2}{3}$.Comparing the constant terms: $1 = A + 2C$ $\Rightarrow 1 = \frac{5}{3} + 2C$ $\Rightarrow 2C = 1 - \frac{5}{3} = -\frac{2}{3}$ $\Rightarrow C = -\frac{1}{3}$.Therefore,$A - B + C = \frac{5}{3} - (-\frac{2}{3}) + (-\frac{1}{3}) = \frac{5+2-1}{3} = \frac{6}{3} = 2$.

If the partial fraction decomposition of $\frac{x^2+1}{x^3+3x^2+3x+2}$ is $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$,then find the value of $A-B+C$. Note: The original expression provided in the prompt was corrected to the standard form $\frac{A}{x+2} + \frac{Bx+C}{x^2+x+1}$.

Similar Questions

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

If $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$,then $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$

If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is given by $A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,then the value of $C$ is

If $\frac{x+1}{(2x-1)(3x+1)}=\frac{A}{2x-1}+\frac{B}{3x+1}$,then $16A+9B$ is equal to

If $\frac{x^2-3x+2}{(x-4)(x-3)^2}=\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{(x-3)^2}$ then $A+B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module