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Question: If \(\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{...

If x(x−1)(x2+1)2=14[1(x−1)−x+1x2+1]+y\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + y then y =

A

(1−x)2(x2+1)2\frac{(1 - x)}{2(x^{2} + 1)^{2}}

B

(1−x)3(x2+1)\frac{(1 - x)}{3(x^{2} + 1)}

C

1+x2(x2−1)2\frac{1 + x}{2(x^{2} - 1)^{2}}

D

None of these

Answer

(1−x)2(x2+1)2\frac{(1 - x)}{2(x^{2} + 1)^{2}}

Explanation

Solution

x(x−1)(x2+1)2=14[1(x−1)−x+1x2+1]+y\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + y

⇒x(x−1)(x2+1)2=14[1(x−1)−x+1x2+1]+Ax+B(x2+1)2\frac{x}{(x - 1)(x^{2} + 1)^{2}} = \frac{1}{4}\left\lbrack \frac{1}{(x - 1)} - \frac{x + 1}{x^{2} + 1} \right\rbrack + \frac{Ax + B}{(x^{2} + 1)^{2}}

⇒4x=(x2+1)2−(x+1)(x−1)(x2+1)+4(Ax+B)(x−1)4x = (x^{2} + 1)^{2} - (x + 1)(x - 1)(x^{2} + 1) + 4(Ax + B)(x - 1)

⇒ 4A+2=04A + 2 = 0,4B−4A=44B - 4A = 4 ⇒ A=−12A = \frac{- 1}{2}, B=12B = \frac{1}{2}

∴\therefore y=Ax+B(x2+1)2=12(1−x)(x2+1)2y = \frac{Ax + B}{(x^{2} + 1)^{2}} = \frac{1}{2}\frac{(1 - x)}{(x^{2} + 1)^{2}}