Express $\ln(3+x)$ and $\frac{1+x}{1-x}$ as Maclaurin series its probably a lot to ask.. but how can I obtain the Maclaurin series for the two functions $f(x)=\ln(3+x)$ and $g(x)=\frac{1+x}{1-x}$ ? as far as I know I cant use any commonly known series to help me with this one ? so finding the derivatives at 0:
$f(0) = \ln(3)$
$f'(x)= 1/(x+3) \implies f'(0) = 1/3$
$f''(x)= -1/(x+3)^2 \implies f''(0)= -1/ 3^2$
$f'''(x)= 2/(x+3)^3 \implies f'''(0)= 2/3^3$
$f''''(x)= -6/(x+3)^4 \implies f''''(0)= -6/3^4$
and now I guess Ill have to use the maclaurin formula: 
$$f(x)=\sum \frac{f^{(n)}(0) x^n}{n!}$$
but now how do I continue from here? I've got:
$$(\ln 3) + \frac{x}{3} - \frac{x^2}{3^2 \cdot 2!} + \frac{2x^3}{3^3 \cdot 3!} - \frac{6x^4}{3^4 \cdot 4!} $$
same question goes for $g(x)$ honestly, on that one upon simplifying the numerator from $g(0)$ with the factorial when I plug it into the maclaurien series formula  I just get $g(x)= 1+2x+2x^2 + 2x^3 + 2x^4 + \cdots$

for $(x+1)/(1-x)$ write it as 
$$
x \frac{1}{1-x} + \frac{1}{1-x}
$$
substitute the series for $1/(1-x)$ and proceed. To get a single series, add the terms one by one. 
Also
$$
\log(3 -x) = \log(3) + \log(1 - x/3)
$$
so substitute $x/3$ in the series for $\log(1-y)$ which can be found be integrating the series for $1/(1-y).$
switch the sign of $x$ to get the series for $\log(3+x).$ Ie multiply the coefficient of $x^n$ by $(-1)^n.$
