find a power series for the given function, centered at a=0.
a)F(x)=ln(3+x)
b) G(x)=1/(1+2x)^(1/4)
c) Integral, 0 to x, (1-cos(t)/t^2) dt
$\displaystyle \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}$.
So $\displaystyle \int{\frac{1}{3 + x}\,dx} = \ln{(3 + x)}$.
So if you have a power series for $\displaystyle \frac{1}{3 + x}$, integrate it, then you will have a power series for $\displaystyle \ln{(3 + x)}$.