How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$

Simply
$$\frac{x}{x+1}=\frac{x+1-1}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=1-\frac{1}{x+1}$$
