Answer
$\frac{\sqrt[4]{x^{2}+1}(2x^{2}+3)}{\sqrt{x}}$
Work Step by Step
We factor out common terms and simplify:
$3x^{-1/2}(x^{2}+1)^{5/4}-x^{3/2}(x^{2}+1)^{1/4}=x^{-1/2}(x^{2}+1)^{1/4}[3(x^{2}+1)-(1)x^{2}]=x^{-1/2}(x^{2}+1)^{1/4}(3x^{2}+3-x^{2})=x^{-1/2}(x^{2}+1)^{1/4}(2x^{2}+3)=\frac{\sqrt[4]{x^{2}+1}(2x^{2}+3)}{\sqrt{x}}$
