Question

\(\dfrac{3}{x-1}-\dfrac{x^3-x}{x^2+1}\left(\dfrac{4}{x^2-2x+1}-\dfrac{4}{x^2-1}\right)\)

Answer 1

\(=\dfrac{3}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{4}{\left(x-1\right)^2}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{3}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{4x+4-4x+4}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{3}{x-1}-\dfrac{x\cdot8}{\left(x^2+1\right)\left(x-1\right)}\)

\(=\dfrac{3x^2+3-8x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{3x^2-8x+3}{\left(x-1\right)\left(x^2+1\right)}\)