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

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

Question

(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[ \frac{1}{(x - 1)} - \frac{x + 1}{x^2 + 1} \right] + y$First,simplify the ter

Vedclass · Accepted Answer

$\frac{1 - x}{2(x^2 + 1)^2}$

Vedclass · Answer

(A) Given the equation: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[ \frac{1}{(x - 1)} - \frac{x + 1}{x^2 + 1} \right] + y$First,simplify the term inside the bracket: $\frac{1}{x - 1} - \frac{x + 1}{x^2 + 1} = \frac{(x^2 + 1) - (x^2 - 1)}{(x - 1)(x^2 + 1)} = \frac{2}{(x - 1)(x^2 + 1)}$Substituting this back: $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4} \cdot \frac{2}{(x - 1)(x^2 + 1)} + y = \frac{1}{2(x - 1)(x^2 + 1)} + y$Solving for $y$: $y = \frac{x}{(x - 1)(x^2 + 1)^2} - \frac{1}{2(x - 1)(x^2 + 1)}$Taking common denominator $2(x - 1)(x^2 + 1)^2$: $y = \frac{2x - (x^2 + 1)}{2(x - 1)(x^2 + 1)^2} = \frac{-(x^2 - 2x + 1)}{2(x - 1)(x^2 + 1)^2} = \frac{-(x - 1)^2}{2(x - 1)(x^2 + 1)^2} = \frac{-(x - 1)}{2(x^2 + 1)^2} = \frac{1 - x}{2(x^2 + 1)^2}$

If $\frac{x}{(x - 1)(x^2 + 1)^2} = \frac{1}{4}\left[ \frac{1}{(x - 1)} - \frac{x + 1}{x^2 + 1} \right] + y$,then $y =$

Similar Questions

What is incorrect for $Nostoc$ and $Anabaena$?

Two sides of a triangle are given by the roots of the equation $x^2-5x+6=0$ and the angle between the sides is $\frac{\pi}{3}$. Then,the perimeter of the triangle is

Which of the following on thermal decomposition produces both basic and acidic oxides?

$H_{3}PO_{3}$ can be represented by structures $1$ and $2$ shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing $H_{3}PO_{3}$? If not,give reasons for the same.

Vedclass Test Series

Exam Paper Generator

Online Exam Module