If frac{x+1}{x^3(x-1)} = frac{a}{x} + frac{b} | vedclass

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

Question

(B) Given the partial fraction decomposition: $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$.Multiplying both si

Vedclass · Accepted Answer

$a = b = 2c = -d$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$.Multiplying both sides by $x^3(x-1)$, we get: $x+1 = ax^2(x-1) + bx(x-1) + c(x-1) + dx^3$.Expanding the right side: $x+1 = a(x^3 - x^2) + b(x^2 - x) + c(x-1) + dx^3$.Grouping the powers of $x$: $x+1 = (a+d)x^3 + (b-a)x^2 + (c-b)x - c$.Comparing coefficients:Constant term: $-c = 1 \implies c = -1$.Coefficient of $x$: $c - b = 1 \implies -1 - b = 1 \implies b = -2$.Coefficient of $x^2$: $b - a = 0 \implies a = b = -2$.Coefficient of $x^3$: $a + d = 0 \implies d = -a = 2$.Checking the values: $a = -2, b = -2, c = -1, d = 2$.We observe that $a = b = 2c = -d$ since $-2 = -2 = 2(-1) = -(2)$.

If $\frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1}$, then:

Similar Questions

Resolve $\frac{x^2 + 1}{(2x - 1)(x^2 - 1)}$ into partial fractions.

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

$\begin{aligned} & \text{If } \frac{4x^2+5x^4+7}{(x^2+1)(x^4+x^2+1)} = \frac{Ax+B}{x^2+1} \\ & + \frac{Cx^3+Dx^2+Ex+F}{x^4+x^2+1}, \text{ then } \\ & B+2(D+F+E)-C \cdot A = \end{aligned}$

If $F_1$ and $F_2$ are irreducible factors of $x^4+x^2+1$ with real coefficients and $\frac{x^3-2x^2+3x-4}{x^4+x^2+1}=\frac{Ax+B}{F_1}+\frac{Cx+D}{F_2}$,then $A+B+C+D=$

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2A-3B+C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module