Answer
$\ln (1+x) = \Sigma_{k=0}^{\infty} \dfrac{(-1)^k x^{k+1}}{k+1}$
Work Step by Step
$\dfrac{1}{1-x}=\Sigma_{k=0}^{\infty} x^k$
We will replace $x$ with $-x$.
$\dfrac{1}{1-(-x)}=\Sigma_{k=0}^{\infty} (-x)^k$
$\dfrac{1}{1+x}=\Sigma_{k=0}^{\infty} (-1)^k (x)^k$
Integrate it.
So, $\ln (1+x) = \Sigma_{k=0}^{\infty} \dfrac{(-1)^k x^{k+1}}{k+1}$
