If frac{x^4+24x^2+28}{(x^2+1)^3} = frac{A}{x^ | vedclass

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

Question

(D) Given the equation: $\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$ Multiply both sides by $(x^2+1)

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Given the equation: $\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$ Multiply both sides by $(x^2+1)^3$: $x^4+24x^2+28 = A(x^2+1)^2 + B(x^2+1) + C$ Let $y = x^2+1$,then $x^2 = y-1$. Substituting this into the equation: $(y-1)^2 + 24(y-1) + 28 = Ay^2 + By + C$ $y^2 - 2y + 1 + 24y - 24 + 28 = Ay^2 + By + C$ $y^2 + 22y + 5 = Ay^2 + By + C$ Comparing the coefficients of $y^2$,$y$,and the constant term: $A = 1$ $B = 22$ $C = 5$ We need to find $A+C$: $A+C = 1 + 5 = 6$

If $\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$,then $A+C=$

Similar Questions

If $\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$,then $2(a+b)=$

What is the quotient when $x^3-5x^2+2x+7$ is divided by $(x-1)$?

The coefficient of ${x^n}$ in the expansion of $\frac{1}{(1 - x)(3 - x)}$ is

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

If $\frac{9}{(x - 1)(x + 2)^2} = \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$,then $A - B - C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module