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} \Bigl(\frac{x^2+1}{x+1}\Bigr)' &= \frac{(x^2+1)'\cdot (x+1) - (x^2+1)\cdot (x+1)'}{(x+1)^2}\\[5pt] &= \frac{2x\cdot (x+1) - (x^2+1)\cdot 1}{(x+1)^2}\\[5pt] &= \frac{2x^2+2x-x^2-1}{(x+1)^2}\\[5pt] &= \frac{x^2+2x-1}{(x+1)^2}\,\textrm{.} \end{align}

Note: It is possible to rewrite the numerator by completing the square,

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

and then the answer can be written as

\displaystyle \frac{x^2+2x-1}{(x+1)^2} = \frac{(x+1)^2-2}{(x+1)^2} = 1-\frac{2}{(x+1)^2}\,\textrm{.}