Answer
$$3\ln \left(2\right)-1$$
Work Step by Step
Given $$ \int_{0}^{1} \ln (4-2 x) d x$$
Let \begin{align*}
u&=\ln (4-2x)\ \ \ \ \ \ \ \ \ \ \ \ \ dv= dx\\
du&= \frac{-1}{2-x}dx\ \ \ \ \ \ \ \ \ \ \ \ v=x
\end{align*}
Then
\begin{align*}
\int_{0}^{1} \ln (4-2 x) d x&=x\ln (4-2x)\bigg|_{0}^{1}+\int_{0}^{1}\frac{xdx}{2-x}\\
&=x\ln (4-2x)\bigg|_{0}^{1}-\int_{0}^{1}\frac{(2-x+2)dx}{2-x}\\
&=x\ln (4-2x)\bigg|_{0}^{1}-\int_{0}^{1}\left(1+\frac{2}{2-x}\right)dx\\
&= x\ln (4-2x) -x+2\ln |2-x| \bigg|_{0}^{1}\\
&= 3\ln \left(2\right)-1
\end{align*}
