Develop the Taylor Series for the function $f(x)={\frac{1}{1+x^2}}-{\frac{1}{x+2}}$ for $x=a$

Question

Develop the Taylor Series for the function $$f(x)={\frac{1}{1+x^2}}-{\frac{1}{x+2}}$$ for $x=a$ and $a=0$ Do I have to take the derivative of both of those fractions? Or for $f_1(x)=\frac{1}{1+x^2}$, can I use the ratio $-x^2$ and use the geometric series resulting this $f_1(x)=\frac{1}{1+x^2}=\sum_{n\ge0}{(-1)^nx^{2n}}$ ? With this being the formula for Taylor Series: $$T_n(x)=\frac{f^{(n)}(x)}{n!}(x-a)$$

Answer 1

The Expansion for $$\frac1{1+x^2}=\frac1{1-(-x^2)}=1-x^2+x^4-x^6...=\sum_{k=0}^\infty \left(-x^2\right)^k$$ The Expansion for $$\frac1{2+x}=\frac12*\frac1{1-(-\frac x2)}=\frac12\left(1-\left(\frac x2\right)+\left(\frac x2\right)^2-\left(\frac x2\right)^3...\right)=\frac12\sum_{k=0}^\infty \left(-\frac x 2\right)^k$$ You can just combine like terms and subtract them.