Aus Online Mathematik Brückenkurs 2

The expression is a quotient of two polynomials, \displaystyle x^{2}+1 and \displaystyle x+1, and we therefore use the quotient rule for differentiation:

\displaystyle \begin{align} & \left( \frac{x^{2}+1}{x+1} \right)^{\prime }=\frac{\left( x^{2}+1 \right)^{\prime }\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot \left( x+1 \right)^{\prime }}{\left( x+1 \right)^{2}} \\ & \\ & =\frac{2x\centerdot \left( x+1 \right)-\left( x^{2}+1 \right)\centerdot 1}{\left( x+1 \right)^{2}} \\ & \\ & =\frac{2x^{2}+2x-x^{2}-1}{\left( x+1 \right)^{2}}=\frac{x^{2}+2x-1}{\left( x+1 \right)^{2}} \\ \end{align}

NOTE: it is possible to rewrite the numerator by completing the square,

\displaystyle x^{2}+2x-1=\left( x+1 \right)^{2}-1^{2}-1=\left( x+1 \right)^{2}-2

and then the answer can be written as

\displaystyle \frac{x^{2}+2x-1}{\left( x+1 \right)^{2}}=\frac{\left( x+1 \right)^{2}-2}{\left( x+1 \right)^{2}}=1-\frac{2}{\left( x+1 \right)^{2}}