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

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

Question

(C) Given the partial fraction decomposition: $\frac{3x + 4}{{(x + 1)}^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^2}$Multiplyi

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(C) Given the partial fraction decomposition: $\frac{3x + 4}{{(x + 1)}^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^2}$Multiplying both sides by the denominator ${(x + 1)}^2(x - 1)$,we get:$3x + 4 = A{(x + 1)}^2 + B(x + 1)(x - 1) + C(x - 1)$To find $A$,substitute $x = 1$ into the equation:$3(1) + 4 = A(1 + 1)^2 + B(1 + 1)(1 - 1) + C(1 - 1)$$7 = A(2)^2 + 0 + 0$$7 = 4A$$A = \frac{7}{4} = 1.75$

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

Similar Questions

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

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

If $\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}$,then $C+D$ is equal to

If $\frac{27x^2+32x+16}{(3x+2)^2(1-x)} = \frac{A}{3x+2} + \frac{B}{(3x+2)^2} + \frac{C}{1-x}$,then $AB+BC+CA =$

If $\frac{13x+43}{2x^2+17x+30} = \frac{A}{2x+5} + \frac{B}{x+6}$,then $A^2+B^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module