The partial fraction decomposition of frac{3x | vedclass

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

Question

(C) We express the fraction as: $\frac{3x+1}{(x+2)(x-1)^2} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$Multiplying both sides by $(x+2)(x-1)^2

Vedclass · Accepted Answer

$\frac{-5}{9} \left(\frac{1}{x+2}\right) + \frac{5}{9} \cdot \frac{1}{x-1} + \frac{4}{3} \cdot \frac{1}{(x-1)^2}$

Vedclass · Answer

(C) We express the fraction as: $\frac{3x+1}{(x+2)(x-1)^2} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$Multiplying both sides by $(x+2)(x-1)^2$,we get:$3x+1 = A(x-1)^2 + B(x-1)(x+2) + C(x+2)$Expanding the terms:$3x+1 = A(x^2 - 2x + 1) + B(x^2 + x - 2) + C(x+2)$$3x+1 = (A+B)x^2 + (-2A + B + C)x + (A - 2B + 2C)$Comparing coefficients:$1$) $A+B = 0 \Rightarrow A = -B$$2$) $-2A + B + C = 3$$3$) $A - 2B + 2C = 1$Substituting $A = -B$ into $(2)$ and $(3)$:$2$) $2B + B + C = 3 \Rightarrow 3B + C = 3$$3$) $-B - 2B + 2C = 1 \Rightarrow -3B + 2C = 1$Adding the two equations: $(3B + C) + (-3B + 2C) = 3 + 1$ $\Rightarrow 3C = 4$ $\Rightarrow C = \frac{4}{3}$Substituting $C = \frac{4}{3}$ into $3B + C = 3$:$3B + \frac{4}{3} = 3$ $\Rightarrow 3B = \frac{5}{3}$ $\Rightarrow B = \frac{5}{9}$Since $A = -B$,$A = -\frac{5}{9}$Thus,the decomposition is $\frac{-5}{9(x+2)} + \frac{5}{9(x-1)} + \frac{4}{3(x-1)^2}$.

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

Similar Questions

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

If the partial fraction decomposition of $\frac{x^4+24x^2+28}{(x^2+1)^3}$ is $\frac{A}{x^2+1}+\frac{B}{(x^2+1)^2}+\frac{C}{(x^2+1)^3}$,then $B-2A+C=$

The partial fraction of $\frac{x^2}{x^2+3x-4}$ is

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module