Question: Simplify: $1 + \dfrac{1}{x-1} +$$\dfrac{2x}{x²+1} -$$\dfrac{x}{x+1} +$$\dfrac{4x³}{x⁴+1}$


Solution:
Given,

$= 1 + \dfrac{1}{x-1} +\dfrac{2x}{x²+1} -\dfrac{x}{x+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{x-1 +1}{x-1} +\dfrac{2x}{x²+1} -\dfrac{x}{x+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{x}{x-1} +\dfrac{2x}{x²+1} -\dfrac{x}{x+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{x}{x-1} -\dfrac{x}{x+1} +\dfrac{2x}{x²+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{x(x+1) -x(x-1)}{(x-1)(x+1)} +\dfrac{2x}{x²+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{x²+x-x²+x}{x²-1} +\dfrac{2x}{x²+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{2x}{x²-1} +\dfrac{2x}{x²+1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{2x(x²+1) +2x(x²-1)}{(x²-1)(x²+1)} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{2x(x²+1+x²-1)}{x⁴-1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{2x(2x²)}{x⁴-1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{4x³}{x⁴-1} +\dfrac{4x³}{x⁴+1}$

$= \dfrac{4x³(x⁴+1) +4x³(x⁴-1)}{(x⁴-1)(x⁴+1)}$

$= \dfrac{4x³(x⁴+1+x⁴-1)}{x⁸-1}$

$= \dfrac{4x³(2x⁴)}{x⁸-1}$

$= \dfrac{8x⁷}{x⁸-1}$
= Answer

Related Notes and Solutions:

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