[tex]\bf y=\cfrac{1}{x+1}+x\qquad \qquad \begin{cases} u=x+1\\ \frac{du}{dx}=1\\[-0.5em] \hrulefill\\ u-1=x \end{cases}\qquad \qquad y=\cfrac{1}{u}+(u-1) \\\\\\ y=u^{-1}+u-1\implies \cfrac{dy}{du}=\stackrel{\textit{chain-rule}}{-u^{-2}\cdot \cfrac{du}{dx}}+\stackrel{\textit{chain-rule}}{1\cdot \cfrac{du}{dx}}+0 \\\\\\ \cfrac{dy}{dx}=-\cfrac{1}{u^2}\cdot 1+1\cdot 1\implies \cfrac{dy}{dx}=-\cfrac{1}{u^2}+1\implies \cfrac{dy}{dx}=-\cfrac{1}{(x+1)^2}+1[/tex]
