Answer
$\frac{4x^3-x^2-x}{(x+1)^{\frac{3}{2}}}$
Work Step by Step
$4x^3(x+1)^{-\frac{3}{2}}-x(x+1)^{-\frac{1}{2}}=(x+1)^{-\frac{3}{2}}[4x^3-x(x+1)^{-\frac{1}{2}-(-\frac{3}{2})}]=\frac{4x^3-x(x+1)^1}{(x+1)^{\frac{3}{2}}}=\frac{4x^3-x^2-x}{(x+1)^{\frac{3}{2}}}$
