How to evaluate this integral:
$$\int\_0^1 \int\_0^1 \cdots \int\_0^1\frac{x\_{1}^2+x\_{2}^2+\cdots+x\_{n}^2}{x\_{1}+x\_{2}+\cdots+x\_{n}}dx\_{1}\, dx\_{2}\cdots \, dx\_{n}=?$$
I'm making use of the integral identity:
$$\int\_{0}^{+\infty }e^{-t(x\_{1}+x\_{2}\cdots +x\_{n})}dt=\frac{1}{x\_{1}+x\_{2}\cdots +x\_{n}}$$ and then reversing the order of integration with respect to time and space variables.
But for $n=1$, then such that, $$\int\_{0}^{\infty }dt\int\_{0}^{1}x^{2}e^{-tx}dx=\int\_{0}^{\infty }\frac{2 - e^{-t}(2 + 2t+t^2)}{t^3}dt=\int\_{0}^{1}x\,dx=\frac{1}{2},$$ and $$\int\_0^1 \int\_0^1 \cdots \int\_0^1\frac{x\_{1}^2+x\_{2}^2+\cdots+x\_{n}^2}{x\_{1}+x\_{2}+\cdots+x\_{n}}dx\_{1}\, dx\_{2}\cdots \, dx\_{n}\\=n\int\_{0}^{+\infty }\frac{2 - e^{-t}\left ( 2 + 2t+t^2 \right )}{t^3}\left ( \frac{1-e^{-t}}{t} \right )^{n-1}dt.$$



Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$
Here is another approach, which also gives the rational term.


(I) To see how it works let $n\geq 2$ and consider first
the simpler case 
\begin{align\*}
\mathbb{E}\bigg(\frac{1}{X\_1+\ldots+X\_n}\bigg)=\int\_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt 
\end{align\*}
Using $\frac{1}{t^n}=\int\_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}\,dz$ we find
\begin{align\*}
\int\_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt&=\int\_0^\infty\int\_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dz\,dt\\
&=\int\_0^\infty\int\_0^\infty \frac{z^{n-1}}{(n-1)!} e^{-zt}(1-e^{-t})^n\, dt\,dz\\
&=\int\_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\
&=n\,\int\_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz
\end{align\*}
The following observation will be the key:


**Lemma**: let $x\_0,\ldots,x\_n$ be distinct positive numbers and $k\leq n-1$. Then
$$I\_{k,n}(x\_0,\ldots,x\_n):=\int\_0^\infty \frac{z^{k}}{(z+x\_0)(z+x\_1)\cdots(z+x\_n)}\,dz=(-1)^{n+k+1}\Delta^n(x^k\log(x);x\_0,\ldots,x\_n)$$
where (for a real function $f$) $\Delta^n(f;x\_0,\ldots,x\_n)$
denotes the divided difference of $f$ corresponding to $x\_0,\ldots,x\_n$.


Recall that (Newton-interpolation)


(1) the divided differences are for $f$ and mutually distinct $x\_0,\ldots,x\_n$ are defined recursively by 
$\Delta^0(f;x\_0)=f(x\_0)$,
$\Delta^n(f;x\_0,\ldots,x\_n)=\frac{\Delta^{n-1}(f; x\_1,\ldots,x\_n)-\Delta^{n-1}(f; x\_0,\ldots,x\_{n-1})}{x\_n-x\_0}$


(2) they are explicitly
given by $$\Delta^n(f;x\_0,\ldots,x\_n)=\sum\_{i=0}^n \frac{f(x\_i)}{\prod\_{j\neq i} (x\_i-x\_j)}\;\;(\*\*)$$
(3) $$\Delta^n(f;x,x+1,\ldots,x+n)=\frac{1}{n!} \sum\_{i=0}^n {n \choose i} (-1)^{n-i} f(x+i)$$


Proof of the lemma:
For $k=0, n=1$ we have $$\int\_0^\infty \frac{1}{(z+x\_0)(z+x\_1)}=\frac{\log(x\_1)-\log(x\_0)}{x\_1-x\_0}=\Delta^1(\log(x);,x\_0,x\_1)$$
For $k=0,n>1$ the repeated use of $\frac{1}{(z+a)(z+b)}=\frac{-1}{b-a}\left(\frac{1}{z+b}-\frac{1}{z+a}\right)$ shows that $(-1)^{n+1}I\_{0,n}(x)=\Delta^n(\log(x),x)$.
The validity for $k>0,n=k+1$ follows from $\frac{z}{z+b}=1-\frac{b}{z+b}$ and $(\*\*)$. End of proof.


The lemma and (3) now give that
\begin{align\*}
 \int\_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt &=n \Delta^{n-1}(x^{n-2}\log(x);x+1,x+2,\ldots,x+n)\\
 &=\frac{n}{(n-1)!}\sum\_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1)
\end{align\*}


(II) Now let $n\geq 1$ and consider
$$ Q\_{n+1}:=\mathbb{E}\bigg(\frac{X\_1^2+\ldots + X\_{n+1}^2}{X\_1+\ldots+X\_{n+1}}\bigg)$$
Write 
$$Q\_{n+1}=(n+1)\int\_0^1 u^2 \bigg(\int\_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt\bigg)\,du$$
 Proceeding as above shows that for $u>0$
\begin{align\*}
 \int\_0^\infty e^{-ut}\bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\frac{n}{n!} \sum\_{i=0}^n {n \choose i} (-1)^{n-i} (u+i)^{n-1}\log(u+i)
\end{align\*}
Thus
$$q\_{n+1}:=\frac{Q\_{n+1}}{n(n+1)}=\frac{1}{n!} \sum\_{i=0}^n {n \choose i} (-1)^{n-i} \int\_0^1 u^2(u+i)^{n-1}\log(u+i)\,du$$
Now expand $u^2=(u+i-i)^2$ and integrate partial to find 
that $q\_{n+1}=L-R$
where $L=\frac{1}{n!}\sum\_{i=0}^n{n\choose i}(-1)^{n-i} \ell(i),\;R=\frac{1}{n!}\sum\_{i=0}^n {n \choose i}(-1)^{n-i} r(i)$ with
\begin{align\*}
 \ell(i)=&\frac{1}{n+2}\left((i+1)^{n+2}\log(i+1)-i^{n+2}\log(i)\right)\\
 &-\frac{2i}{n+1}\left((i+1)^{n+1}\log(i+1)-i^{n+1}\log(i)\right)\\
 &+\frac{i^2}{n}\left((i+1)^{n}\log(i+1)-i^n\log(i)\right)\\[0.2cm]
 r(i)=&\frac{1}{(n+2)^2}\left((i+1)^{n+2}-i^{n+2}\right)\\
 &-\frac{2i}{(n+1)^2}\left((i+1)^{n+1}-i^{n+1}\right)\\
 &+\frac{i^2}{n^2}\left((i+1)^{n}-i^n\right)
\end{align\*}
Collecting terms in the logarithmic part $L$ shows that the coefficient of $\log(i)$ in $q\_{n+1}$ is $c\_{i,n+1} =(-1)^{n-i} \frac{i^n}{n}{ n \choose i-1}\left(\frac{n^2+3n+2-2i}{(n+2)!}\right)$, matching Fedor Petrov's answer.


The rational part $R$ can readily be summed (note that only the contributions of the powers $n$ and $n+1$ (with coefficients $-\frac{1}{n(n+1)(n+2)}$ and $\frac{2}{n(n+1)(n+2)}$) need to be considered) to
give $$R=\frac{1}{n+2}-\frac{1}{n(n+1)(n+2)}\;\;,$$
so that the rational term of $Q\_{n+1}$ is $-\frac{n(n+1)-1}{n+2}=-\frac{(n+2)(n-1)+1}{n+2}$, confirming Sylvain JULIEN's guess.


