(a) Simplify $\dfrac{( x^{2}+1) ( 3) -( 3x+4) ( 2x) }{( x^{2}+1) ^{2}}$.

(b) Write the expression $( x^{2}+1) ^{1/2}+x\cdot \dfrac{1}{2}( x^{2}+1) ^{-1/2}\cdot 2x$ as a single quotient in which only positive exponents appear.

Solution (a) $$\begin{eqnarray*} \ \dfrac{( x^{2}+1) (3) -( 3x+4) ( 2x) }{( x^{2}+1) ^{2}}&=&\dfrac{ 3x^{2}+3- ( 6x^{2}+8x) }{( x^{2}+1) ^{2}}=\dfrac{ 3x^{2}+3-6x^{2}-8x}{( x^{2}+1) ^{2}}\\[6pt] &=&\dfrac{-3x^{2}-8x+3}{( x^{2}+1) ^{2}}\underset{\underset{\color{#0066A7}{\hbox{Factor}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{- ( 3x-1) ( x+3) }{( x^{2}+1) ^{2}}\\ \end{eqnarray*}$$

(b) $$\begin{eqnarray*} ( x^{2}+1) ^{1/2}+x\cdot \dfrac{1}{2}( x^{2}+1) ^{-1/2}\cdot 2x&=&( x^{2}+1) ^{1/2}+\dfrac{x^{2}}{ ( x^{2}+1) ^{1/2}}\\[6pt] &=&\dfrac{( x^{2}+1) ^{1/2}( x^{2}+1) ^{1/2}}{( x^{2}+1) ^{1/2}}+\dfrac{x^{2}}{( x^{2}+1) ^{1/2}}\\[6pt] &=&\dfrac{( x^{2}+1) +x^{2}}{( x^{2}+1) ^{1/2}}=\dfrac{ 2x^{2}+1}{( x^{2}+1) ^{1/2}} \end{eqnarray*}$$