Answer
$\Sigma_{k=0}^{n} \dfrac {(-1)^kx^{k+1}}{k+1}$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\ln (1+x)$
Differentiate w.r.t x
$f'(x)=\dfrac{1}{(1+x)}$
$f''(x)=-\dfrac{1}{(1+x)^2}$
$f'''(x)=\dfrac{2}{(1+x)^3}$
$f''''(x)=\dfrac{-6}{(1+x)^4}$
Now $f(0)=0$, $f'(0)=1$, $f''(0)=-1$, $f'''(0)=2$ and $f''''(0)=-6$
Hence the Maclaurin Series for the function is
$\displaystyle x - \frac {x^2} {2} + .......+ \frac {(-1)^nx^{n+1}}{n+1} = \Sigma_{k=0}^{n} \frac {(-1)^kx^{k+1}}{k+1}$
