## College Algebra 7th Edition

$\dfrac{x+1}{(x^2+1)(x-1)}$
The LCD of the rational expressions is $(x-1)(x^2+1)$. Make the expressions similar using their LCD to obtain: $=\dfrac{1(x^2+1)}{(x-1)(x^2+1)} - \dfrac{x(x-1)}{(x^2+1)(x-1)} \\=\dfrac{1(x^2)+1(1)}{(x-1)(x^2+1)}- \dfrac{x(x) - 1(x)}{(x^2+1)(x-1)} \\=\dfrac{x^2+1}{(x-1)(x^2+1)}-\dfrac{x^2-x}{(x^2+1)(x-1)}$ Subtract the numerators together and copy the denominator to obtain: $=\dfrac{x^2+1 -(x^2-x)}{(x^2+1)(x-1)} \\=\dfrac{x^2+1 -x^2-(-x)}{(x^2+1)(x-1)} \\=\dfrac{x^2+1 -x^2+x}{(x^2+1)(x-1)} \\=\dfrac{(x^2-x^2)+x+1}{(x^2+1)(x-1)} \\=\dfrac{x+1}{(x^2+1)(x-1)}$