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

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

Question

(NONE) Given $\frac{x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by $(x-1)^2(x^2+1)$,we get $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(NONE) Given $\frac{x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$.Multiplying both sides by $(x-1)^2(x^2+1)$,we get $x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$.For $x=1$,$1+1 = B(1^2+1) \implies 2 = 2B \implies B=1$.Comparing coefficients of $x^3$: $0 = A+C \implies C = -A$.Comparing coefficients of $x^0$: $1 = -A + B + D \implies 1 = -A + 1 + D \implies D = A$.Comparing coefficients of $x^2$: $0 = -A + B + C - 2D \implies 0 = -A + 1 - A - 2A \implies 4A = 1 \implies A = \frac{1}{4}$.Thus,$A = \frac{1}{4}$,$B = 1$,$C = -\frac{1}{4}$,$D = \frac{1}{4}$.Now,$\sqrt{3A^2+4D^2+5C^2+B^2} = \sqrt{3(\frac{1}{16}) + 4(\frac{1}{16}) + 5(\frac{1}{16}) + 1} = \sqrt{\frac{3+4+5}{16} + 1} = \sqrt{\frac{12}{16} + 1} = \sqrt{\frac{3}{4} + 1} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}$.

If $\frac{x+1}{(x-1)^2(x^2+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}$,then $\sqrt{3A^2+4D^2+5C^2+B^2}=$

Similar Questions

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

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$

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

$\frac{4x^2+5}{(x-2)^4} = \frac{A}{(x-2)} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4}$,then $\sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = $

Resolve $\frac{x + 1}{(x - 1)(x - 2)(x - 3)}$ into partial fractions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module