Note that when the constant in the denominator is not $1,$ we still can use the geometric series, but we have to multiply by a form of $1$ that helps us$.$ We'll use $\frac{\frac{1}{8}}{\frac{1}{8}}.$

$\frac{1}{(8+x)}=\frac{\frac{1}{8}}{\frac{1}{8}(8+x)}=\frac{\frac{1}{8}}{1+\frac{1}{8}x}=\frac{1}{8}*\frac{1}{1+\frac{1}{8}x}$

And now we can use the geometric series expression to get

$\frac{1}{(8+x)}=\frac{1}{8}*\frac{1}{1+\frac{1}{8}x}=\frac{1}{8}{\displaystyle \underset{n=0}{\overset{}{}}}(\frac{-1}{8}x{)}^n=\frac{1}{8}{\displaystyle \underset{n=0}{\overset{}{}}}(\frac{-1}{8}{)}^n{x}^n$

Use differentiation and$/$or integration to express the following function as a power series $($centered at $x=0).$

$f(x)={\displaystyle \underset{n=0}{\overset{}{}}},f(x)=\frac{1}{(8+x{)}^2}$