<s>[INST] 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??
 [/INST] 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:
$$\frac{1}{(n-1)!}\sum_{i=0}^n(-1)^{n+i}{n \choose i}(a+i)^n\ln(a+i)$$
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