If frac{5x^2+2}{x^3+x}=frac{A_1}{x}+frac{A_2x | vedclass

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

Question

(C) Given that,$\frac{5x^2+2}{x(x^2+1)} = \frac{A_1}{x} + \frac{A_2x+A_3}{x^2+1}$ Multiplying both sides by $x(x^2+1)$,we get: $5x^2+2 = A_1(x^2+1) +

Vedclass · Accepted Answer

$(2, 3, 0)$

Vedclass · Answer

(C) Given that,$\frac{5x^2+2}{x(x^2+1)} = \frac{A_1}{x} + \frac{A_2x+A_3}{x^2+1}$ Multiplying both sides by $x(x^2+1)$,we get: $5x^2+2 = A_1(x^2+1) + (A_2x+A_3)x$ $5x^2+2 = A_1x^2 + A_1 + A_2x^2 + A_3x$ $5x^2+2 = (A_1+A_2)x^2 + A_3x + A_1$ Comparing the coefficients of $x^2$,$x$,and the constant term on both sides: Constant term: $A_1 = 2$ Coefficient of $x$: $A_3 = 0$ Coefficient of $x^2$: $A_1 + A_2 = 5$ $\Rightarrow 2 + A_2 = 5$ $\Rightarrow A_2 = 3$ Thus,$(A_1, A_2, A_3) = (2, 3, 0)$.

If $\frac{5x^2+2}{x^3+x}=\frac{A_1}{x}+\frac{A_2x+A_3}{x^2+1}$,then $(A_1, A_2, A_3) = $

Similar Questions

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

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

If $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$ and $\frac{x+1}{(x-1)^2(x^2+x+1)} = A f_1(x) + (B + \frac{D}{x-1}) f_2(x) + \frac{C}{(x-1)^2}$,then find $A+B+C+D$.

If $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$,then

If $\frac{42-13x}{x^2+x-6}=\frac{A}{lx+m}+\frac{B}{px+q}$ where $lm > 0$ and $pq < 0$,then $\frac{Alp}{Bmq} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module