No idea what the previous part of the problem is, but you have
[tex]f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n[/tex]
[tex]f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}[/tex]
which is valid for [tex]\left|-\dfrac{x^2}4\right|<1[/tex], or [tex]|x|<2[/tex]. So the integral from 0 to 2 is
[tex]\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx[/tex]
[tex]=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx[/tex]
Note that since the power series only converges on the interval if [tex]x[/tex] is strictly less than 2, which means we have to treat this as an improper integral.
[tex]=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx[/tex][/tex]
[tex]=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}[/tex]
[tex]=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}[/tex]
[tex]=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}[/tex]
[tex]=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots[/tex]
