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 the second function, rewrite it as
$$ g(x) = -1 + \frac{2}{1-x} $$
Then apply what you know for $\frac{1}{1-x}$, which is the geometric series
$$ -1 + 2\cdot\frac{1}{1-x} = -1 + 2\sum_{n=0}^{\infty}x^n $$
The first function is a bit tricky, note that
$$\frac{d}{dx}\ln(x+3) = \frac{1}{x+3}$$
With that in mind, we can find the series expression for the derivative and then integrate the series
$$\frac{1}{x+3} = \frac{1}{3\left(1 + \frac{x}{3}\right)} = \frac{1}{3}\cdot\frac{1}{1 - \left(-\frac{x}{3}\right)} \\= \frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{x}{3}\right)^n = \sum_{n=0}^\infty \frac{(-1)^n x^n}{3^{n+1}}$$
Integrating
$$\begin{align}\ln(x+3) &= \int \frac{1}{x+3}\,dx \\&= \int \sum_{n=0}^\infty \frac{(-1)^n x^n}{3^{n+1}} dx \\&= \sum_{n=0}^\infty \int \frac{(-1)^n x^n}{3^{n+1}}\,dx \\&= \sum_{n=0}^{\infty} \frac{(-1)^nx^{n+1}}{(n+1)3^{n+1}} \\&= \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n\cdot3^n} \end{align}$$
