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

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

Question

(B) Let $y = x^2$. The expression becomes $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$.Using p

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) Let $y = x^2$. The expression becomes $\frac{y}{(y+2)(y^2-1)} = \frac{y}{(y+2)(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1} + \frac{C}{y+2}$.Using partial fraction decomposition:$y = A(y+1)(y+2) + B(y-1)(y+2) + C(y-1)(y+1)$.For $y=1$: $1 = A(2)(3) \implies 6A = 1 \implies A = \frac{1}{6}$.For $y=-1$: $-1 = B(-2)(1) \implies -2B = -1 \implies B = \frac{1}{2}$.For $y=-2$: $-2 = C(-3)(-1) \implies 3C = -2 \implies C = -\frac{2}{3}$.Thus,$A+B-C = \frac{1}{6} + \frac{1}{2} - (-\frac{2}{3}) = \frac{1}{6} + \frac{3}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3}$.

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

Similar Questions

If $\frac{x^{4}}{(x - 1)(x - 2)} = f(x) + \frac{A}{x - 1} + \frac{B}{x - 2}$,then

The partial fraction of $\frac{x^2}{(x - 1)^3(x - 2)}$ is

If $\frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b}$,then $a:b$ is equal to

If $\frac{3x^2 + 5}{(x^2 + 1)^2} = \frac{a}{x^2 + 1} + \frac{b}{(x^2 + 1)^2}$,then $(a, b) = $

If $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A x+B}{x^2+1}+\frac{C x+D}{\left(x^2+1\right)^2}+\frac{E x+F}{\left(x^2+1\right)^3}$,then the value of $A+B+C+D+E+F=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module