Explicit Formula for Multiple Integral $$ \int_0^1 \! \int_0^1 ...\int_0^1 \frac{dx_1dx_2...dx_n}{(a+x_1+x_2+...+x_n)^m}$$ 
I've tried for 
n=2, m=1, I got : $a\ln a - 2(a+1)\ln (a+1)+(a+2)\ln(a+2)$
n=2, m=2, I got : $-\ln a + 2\ln(a+1)-\ln(a+2)$
n=3, m=1, I got : $\frac{1}{2}(-a^2\ln a+3(a+1)^2\ln (a+1)-3(a+2)^2\ln(a+2)+(a+3)^2\ln (a+3))$
but I still have no idea what's next. Or there is a tricky way??
 A: Let
$$
f_{n,m}(a)=\int_0^1\cdots\int_0^1\frac{\prod_n dx_i}{(a+\sum_n x_i)^m}
$$
Now, it is easy enough to confirm that
$$
\frac{df_{n,m}}{da}=-mf_{n,m+1}(a)
$$
It is also easy enough to see that
$$
f_{n+1,m}(a)=\int_0^1 f_{n,m}(x+a)\ dx \tag{1}
$$
which, on taking the derivative, gives
$$
f_{n+1,m}'(a) = f_{n,m}(a+1)-f_{n,m}(a)
$$
Bringing these together gives
$$
-mf_{n+1,m+1}(a) = f_{n,m}(a+1)-f_{n,m}(a) = \Delta f_{n,m}(a)
$$
where $\Delta$ refers to the forward difference, or
$$
f_{n+1,m+1}(a)=-\frac{\Delta f_{n,m}(a)}{m}
$$
Iterating down, we can see that, for non-integer $m$, and also for integer $m\geq n$,
$$
f_{n,m}(a)=(-1)^{n-1}\frac{\Gamma(m-n+1)}{\Gamma(m)}\Delta^{n-1} f_{1,m-n+1}(a)
$$
Now, we can calculate for $k\neq1$,
$$
f_{1,k}(a)=\int_0^1 \frac{dx}{(a+x)^k} = \frac1{k-1}\left(\frac1{a^{k-1}}-\frac1{(a+1)^{k-1}}\right) = \frac{\Delta(a^{1-k})}{1-k}
$$
and
$$
f_{1,1}(a)=\log(a+1)-\log(a)
$$
To deal with integer $m\leq n$, we make use of equation (1), and it can be confirmed by induction that
$$
f_{n,1}(a) = \frac{\Delta^n (a^{n-1}\log(a))}{\Gamma(n)} \tag{2}
$$
and we can then write
$$
f_{n,m}(a)=(-1)^{m-1}\frac{\Delta^{m-1}f_{n-m+1,1}(a)}{\Gamma(m)}
$$
and so, the general form of the solution is
$$
f_{n,m}(a)=\begin{cases}
\frac{(-1)^{m-1}}{\Gamma(n-m+1)\Gamma(m)}\Delta^{n} (a^{n-m}\log(a)) &n\geq m\in\mathbb{Z}^+\\
(-1)^n\frac{\Gamma(m-n)}{\Gamma(m)}\Delta^n (a^{n-m}) & \text{otherwise}
\end{cases}
$$
where for integer $m\leq0$, the ratio of gamma functions should be evaluated as one would a ratio of negative factorials.

The Induction (how to get/confirm equation (2)):
The key to this equation is that, when you apply equation (1), the polynomial terms get cancelled out by the repeated Forward Difference operator application.
I hypothesised the form of the equation by examining the first few terms (first one as seen above, second and third both provided in question):
$$\begin{align}
f_{1,1}(a)&=\log(a+1)-\log(a)=\Delta(\log(a))\\
f_{2,1}(a)&=(a+2)\log(a+2)-2(a+1)\log(a+1)+a\log(a) = \Delta^2(a\log(a))\\
f_{3,1}(a)&=\frac{(a\!+\!3)^2\log(a\!+\!3)-3(a\!+\!2)^2\log(a\!+\!2)+3(a\!+\!1)^2\log(a\!+\!1)-a^2\log(a)}{2}\\
&=\frac{\Delta^3(a^2\log(a))}{2}
\end{align}$$
Each power of $a$ in front of the $\log$ terms is one less than $n$, the $\Delta$ iteration matches $n$, and the denominator appears to be following a factorial pattern (1,1,2) - checking $n=4$ (see Dr. Wolfgang Hintze's answer for its expression) can help to confirm the factorial pattern, as the denominator is 6.
So we hypothesise
$$
f_{n,1}(a)=\frac{\Delta^n(a^{n-1}\log(a))}{(n-1)!}=\frac{\Delta^n(a^{n-1}\log(a))}{\Gamma(n)}
$$
where it has been expressed with $\Gamma$ in order to match the form used for other cases.
Now for the induction to confirm. We can see that it's right for $n=1$ easily, so I'll focus on the induction step. Assume $f_{n,1}(a)$ fits that form - can we confirm that $f_{n+1,1}(a)$ also fits that form?
Using equation (1), and noting that the Forward Difference operator commutes with integration,
$$\begin{align}
f_{n+1,1}(a)&=\int_0^1 f_{n,1}(x+a)\ dx\\
&=\frac{\Delta^n}{\Gamma(n)}\int_0^1 (x+a)^{n-1}\log(x+a)\ dx
\end{align}$$
We apply integration by parts with $u=\log(x+a)$ and $dv=(x+a)^{n-1}dx$ to get
$$\begin{align}
\int_0^1 (x+a)^{n-1}\log(x+a)\ d&x=\left[\frac{(x+a)^n}n\log(x+a)\right]_0^1-\int_0^1 \frac{(x+a)^{n-1}}ndx\\
&=\frac{\Delta(a^n\log(a))}n - \left[\frac{(x+a)^n}{n^2}\right]_0^1\\
&=\frac{\Delta(a^n\log(a))}n - \frac{\Delta(a^n)}{n^2}
\end{align}$$
And so we have
$$\begin{align}
f_{n+1,1}(a)&=\frac{\Delta^n}{\Gamma(n)}\left(\frac{\Delta(a^n\log(a))}n - \frac{\Delta(a^n)}{n^2}\right)\\
&=\frac{1}{n\Gamma(n)}\left(\Delta^{n+1}(a^n\log(a)) - \frac{\Delta^{n+1}(a^n)}{n}\right)
\end{align}$$
Now, it can be easily confirmed that $\Delta^n(a^m)=0$ if $n>m\geq0$ and $m$ is an integer (because $\Delta(a^m)$ is a polynomial of order $m-1$). Also, $\Gamma(n+1)=n\Gamma(n)$. So we have
$$
f_{n+1,1}(a)=\frac{\Delta^{n+1}(a^n\log(a))}{\Gamma(n+1)}
$$
which is what we needed to show. By induction, this proves our formula.
A: Use the integral representation
$$
\frac{1}{(a+y)^m}=\frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-\xi (a+y)}
$$
to rewrite your integral as
$$
I(m,n)=\int_0^1 \! \int_0^1 ...\int_0^1 \frac{dx_1dx_2...dx_n}{(a+x_1+x_2+...+x_n)^m}=
\frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-a \xi}\left[\int_0^1 dx  e^{-\xi x}\right]^n
$$
$$
=\frac{1}{\Gamma(m)}\int_0^\infty d\xi \xi^{m-1}e^{-a \xi}\left[\frac{1-e^{-\xi }}{\xi }\right]^n\ .
$$
Expanding via the binomial theorem
$$
I(m,n)=\frac{1}{\Gamma(m)}\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_0^\infty d\xi\ \xi^{m-1-n}e^{-\xi (a+n-k)}
=\frac{1}{\Gamma(m)}\sum_{k=0}^n {n\choose k}(-1)^{n-k}\Gamma (m-n) (a-k+n)^{n-m}\ .
$$
For certain combinations of $n$ and $m$, you'll need to evaluate the expression as a limit, for example for $n=2$ and $m=1$, you will need to evaluate $\lim_{m\to 1} I(m,2)$, which coincides with the value found above.
A: Not a Complete Answer
For $n=4$, $m=1$ Mathematica gave me a similar answer:
$$\frac{1}{6}\left(a^3\ln a-4(a+1)^3\ln(a+1)+6(a+2)^3\ln(a+2)-4(a+3)^3\ln(a+3)+(a+4)^3\ln(a+4)\right)$$
And for $n=5$, $m=1$ it gave me:
$$\frac{1}{24}\left(-a^4\ln a+5(a+1)^4\ln(a+1)-10(a+2)^4\ln(a+2)+10(a+3)^4\ln(a+3)+-5(a+4)^4\ln(a+4)+(a+5)^4\ln(a+3)\right)$$
So for the $m=1$ family it may be reasonable to conclude that the answer is:
$$\frac{1}{(n-1)!}\sum_{i=0}^n(-1)^{n+i}{n \choose i}(a+i)^n\ln(a+i)$$
