Find $\log (2-x)$ in powers of x Find $\log (2-x)$ in powers of $x$. 
I know this is how I've gotten so fare, 
$$\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$$
given that $\left| x\right| <1$.
Now
$$\begin{align}\frac{1}{2-x}&=\frac{1}{2} \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n \end{align}$$
I therefore try integrating
$$\begin{align} \int \frac{1}{2-x} \, dx&=-\log(2 - x)\\
\int \frac{1}{2} \left(\frac{x}{2}\right)^n dx&=
\frac{2^{-n-1} x^{n+1}}{n+1}\end{align}$$
Now multiplying both sides with $-1$ i still not get the answer which is 
$$\begin{align}\ln(2-x)&=\ln(2)-\sum_{n=1}^{+\infty}\frac{x^n}{2^nn} & -2\leq x<2\end{align}$$
Could someone please show me the steps?

We have that $$\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$$
given that $\left| x\right| <1$.
Since $$\dfrac{d}{dx} \log(2-x) =-\dfrac{1}{2-x} $$
we have to transform $-\dfrac{1}{2-x}$ somehow into $\dfrac{1}{1-x}$. 
$$-\dfrac{1}{2-x}=-\dfrac{1}{2(1-\frac{x}{2})}=-\dfrac{1}{2}\cdot \dfrac{1}{1-\frac{x}{2}}= -\dfrac{1}{2}\cdot \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$
Because the series $$\sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$ is uniformly convergent for $|\frac{x}{2}|<1 \Leftrightarrow x\in (-2,2)$ we can integrate the following
$$-\dfrac{1}{2-x}=-\dfrac{1}{2}\cdot \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$
and we'll get that
$$ \int_0^x \left(-\dfrac{1}{2-t}\right)dt = -\dfrac{1}{2}\cdot \sum _{n=0}^{\infty }\int_0^x \left(\frac{t}{2}\right)^n dt $$
After the integration we get:
$$\log(2-x)-\log2=-\sum _{n=0}^{\infty } \dfrac{x^{n+1}}{(n+1)\cdot 2^{n+1}}$$
Let's say that $k=n+1$ then we have that:
$$\log(2-x)=\log2-\sum _{k=1}^{\infty } \dfrac{x^{k}}{k\cdot 2^{k}}$$
Or
$$\log(2-x)=\log2-\sum _{n=1}^{\infty } \dfrac{x^{n}}{n\cdot 2^{n}}$$
