$\text{c)} \dfrac{x^4-x^3+4x^2-x+5}{x^2+1}$
=$\dfrac{(x^4-1)-(x^3+x)+(4x^2+4)+2}{x^2+1}$
=$\dfrac{(x^2+1)(x^2-1)x(x^2+1)+4(x^2+1)+2}{x^2+1}$
=$\dfrac{(x^2+1)(x^2-1-x+4)+2}{x^2+1}$
=$x^2-x+3+\dfrac{2}{x^2+1}$
$\text{d)}\dfrac{x^5-2x^4-x-3}{x+1}$
=$\dfrac{(x^5+x^4)-(x^4-1)-x-4-2x^4}{x+1}$
=$\dfrac{x^4(x+1)-(x^2+1)(x-1)(x+1)-(x+1)-(2x^4-2)-5}{x+1}$
=$\dfrac{x^4(x+1)-(x^2+1)(x-1)(x+1)-(x+1)-2(x^2+1)(x-1)(x+1)-5}{x+1}$
=$\dfrac{(x+1)[x^4-(x^2+1)(x-1)-1-2(x^2+1)(x-1)]-5}{x+1}$
=$x^4-x^3-x^2+x-1+1-2x^3-2x^2+2x+2-\dfrac{5}{x+1}$
=$x^4-3x^3-3x^2+3x+2-\dfrac{5}{x+1}$