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?

Consider the series
\begin{align}
\frac{1}{2-x} = \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n}} = 1 + \sum_{n=1}^{\infty} \frac{x^{n}}{2^{n}}
\end{align}
Now this series can be integrated from 0 to $x$ to obtain
\begin{align}
\int_{0}^{x} \frac{dt}{2-t} &= \int_{0}^{x} \left( 1 + \sum_{n=1}^{\infty} \frac{ t^{n}}{2^{n}} \right) \, dt \\
\left[ - \ln(2-t) \right]_{0}^{x} &= \left[ t + \sum_{n=1}^{\infty} \frac{t^{n+1}}{2^{n} (n+1)} \right]_{0}^{x} \\
- \ln(2-x) + \ln(2) &= x + \sum_{n=1}^{\infty} \frac{x^{n+1}}{2^{n}(n+1)} 
\end{align}
The series can now be changed slightly and is seen as:
\begin{align}
\ln(2-x) &= \ln(2) - x - \sum_{n=2}^{\infty} \frac{x^{n}}{2^{n-1} \, n} \\
&= \ln(2) - \sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} \, n}
\end{align}
