The integral $\int_0^1 x^n \log(2-x) \, dx$ In this question I had stuck on the integral:
$$\int_0^1 x^n \log(2-x) \, {\rm d}x$$
Claude Leibovici says that $\displaystyle \int_0^1 x^n \log(2-x) \, {\rm d}x=\frac{\, _2F_1\left(1,n+2;n+3;\frac{1}{2}\right)}{2 (n+1) (n+2)}=a_n+b_n \log 2$ where $a_n, b_n$ are sequences of rational numbers. The statement seems to be true since if we run the integral for some values of $n$ we get that:
$$\begin{array}{||c|c|c|c|c||}
\hline
\text{Integral}&n & a_n & b_n & \text{Value} \\
\hline
\int_0^1 x \log(2-x) \, {\rm d}x&1 & -\frac{5}{4} & 2 & -\frac{5}{4}+ 2 \log 2 \\\\
\int_0^1 x^2 \log(2-x) \, {\rm d}x&2 & -\frac{16}{9} & \frac{8}{3} & -\frac{16}{9} + \frac{8}{3} \log 2 \\\\
\int_0^1 x^3 \log(2-x) \, {\rm d}x&3 & -\frac{131}{48} & 4& -\frac{131}{48} + 4 \log 2 \\
\hline
\end{array}$$
It seems that $a_n$ is a sequence of negative rational numbers and $b_n$ a sequence of positive rational numbers. However I don't see a pattern of how to connect those numbers. Does anyone see? 
Maybe it is worth it to take a look at the more general case:
$$\int_0^1 x^n \log(\alpha -x) \, {\rm d}x \; \quad \mathbb{N} \ni \alpha \geq 2$$
That would be more tedious. What I had come as a solution but I could not see how to proceed was something like this:
\begin{align*}
\int_{0}^{1}x^n \log(\alpha -x) \, {\rm d}x &= \int_{0}^{1}x^n \log \left [ \alpha \left ( 1 - \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\ 
 &=\int_{0}^{1}x^n \left [ \log \alpha + \log \left ( 1- \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\ 
 &=\log \alpha \int_{0}^{1} x^n \, {\rm d}x + \int_{0}^{1} x^n \log \left ( 1- \frac{x}{\alpha} \right ) \, {\rm d}x \\ 
 &= \frac{\log \alpha}{n+1} - \int_{0}^{1} x^n \sum_{n=1}^{\infty} \frac{\left ( \frac{x}{\alpha} \right )^n}{n} \, {\rm d}x \\ 
 &= \frac{\log \alpha}{n+1} - \sum_{n=1}^{\infty} \frac{1}{n \alpha^n} \int_{0}^{1} x^{2n} \, {\rm d}x\\ 
 &= \frac{\log \alpha}{n+1} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} \\
 &= ??
\end{align*}
It seems that:
$$\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} = -2\sqrt{\alpha} \; {\rm arctanh}  \left ( \frac{1}{\sqrt{\alpha}} \right )  - \log \left ( \frac{\alpha -1}{\alpha} \right ) +2$$
I guess that won't be too difficult to prove taking into account that:
$${\rm arctanh}(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} $$
 A: We can easily develop a recursive formula for the integral of interest.  To that end we proceed.
Let $I_n(\alpha)=\int_0^1 x^n \log(\alpha -x)\,dx$.  Integrating by parts with $u=x^n$ and $v=(x-\alpha)\log(\alpha -x)-x$ yields
$$\begin{align}
I_n(\alpha)&=(1-\alpha)\log(\alpha-1)-1-n\int_0^1 x^{n-1}\left((x-\alpha)\log(\alpha -x)-x\right)\,dx\\\\
&=(1-\alpha)\log(\alpha-1)-nI_n(\alpha)+n\alpha I_{n-1}(\alpha)-\frac{1}{n+1}\\\\
&=\frac{(1-\alpha)\log(\alpha-1)+n\alpha I_{n-1}(\alpha)-\frac{1}{n+1}}{n+1}
\end{align}$$
Therefore, we find that
$$\bbox[5px,border:2px solid #C0A000]{I_n(\alpha)=\frac{(1-\alpha)\log(1-\alpha)}{n+1}-\frac{1}{(n+1)^2}-\frac{2n}{n+1}I_{n-1}(\alpha)}$$
For $\alpha =2$ this yields
$$\bbox[5px,border:2px solid #C0A000]{I_n(2)=\frac{2n}{n+1}I_{n-1}(2)-\frac{1}{(n+1)^2}}$$
A: By integration by parts
$$ I_n=\int_{0}^{1}x^n\log(2-x)\,dx = \frac{1}{n+1}\int_{0}^{1}\frac{x^{n+1}}{2-x}\,dx \tag{1}$$
so it is enough to write $x^{n+1}$ as $(x^{n+1}-2^{n+1})+2^{n+1}$ to get:
$$\begin{eqnarray*} I_n &=& \frac{1}{n+1}\left(2^{n+1}\log 2-\int_{0}^{1}\sum_{k=0}^{n}2^k x^{n-k}\,dx\right)\\&=&\color{blue}{\frac{2^{n+1}}{(n+1)}\log 2}-\color{red}{\frac{1}{n+1}\sum_{k=0}^{n}\frac{2^k}{n-k+1}}.\tag{2}\end{eqnarray*} $$
