Proving a symmetric Cauchy matrix is positive semidefinite
Hint (without integrals): Let $X=\operatorname{diag}(x_1,x_2,\ldots,x_n)$ and $e$ be the vector of all ones.
- Prove that the Cauchy matrix $C$ satisfies the equation $$ XC+CX=ee^T. $$
- For any eigenvalue $\lambda$ of $C$ with $Cv=\lambda v$, pre-multiply the equation by $v^T$ and post-multiply by $v$. Conclude that $\lambda\ge 0$.
Regarding the hint: just as a sum of positive semidefinite operators is positive semidefinite, so is an integral of positive semidefinite operators positive semidefinite. So, since $T(t)$ is positive semidefinite for all $t \geq 0$, it follows that $$ A = \int_0^1 T(t)\,dt $$ is positive semidefinite.
This paper hints at an interesting proof (without integrals) that when the numbers $x_i$ are distinct, $A$ is necessarily positive definite. We note that the determinant of a Cauchy matrix $x_i$ is given by $$ \det(A) = \frac{\prod_{i,k,i>k}(x_i-x_k)^2}{\prod_{i,j = 1}^n(x_i+x_j)}. $$ Because all terms being multiplied are positive, $\det(A) > 0$. Since every principal submatrix of $A$ is itself a Cauchy matrix for some set of distinct $x_i$, we can conclude that the principal submatrices of $A$ also have positive determinant. By Sylvester's criterion, we can conclude that $A$ is positive definite.
For the more general statement with non-distinct $x_i$, it suffices to note that the limit of a sequence positive semidefinite matrices must itself be positive semidefinite.
There is also a proof that is similar in spirit to the hint you've mentioned. See exercise 1.6.3 (pp.24-25) of Bhatia, Positive Definite Matrices. The idea is that, instead of writing the Cauchy matrix as an integral of Gramians, we write it as an infinite sum of Gramians. More specifically, let $0<t<\min_ix_i$. Then $$ \frac{1}{x_i+x_j-t} =\frac{t}{x_ix_j}\left(\frac{1}{1-\frac{(x_i-t)(x_j-t)}{x_ix_j}}\right) =\frac{t}{x_ix_j}\sum_{k=0}^\infty\left(\frac{(x_i-t)(x_j-t)}{x_ix_j}\right)^k. $$ Therefore, by setting $\mathbf v_k=\left(\frac{(x_1-t)^k}{x_1^{k+1}},\,\frac{(x_2-t)^k}{x_2^{k+1}},\,\ldots,\,\frac{(x_n-t)^k}{x_n^{k+1}}\right)^\top$, we see that $$ \left(\frac{1}{x_i+x_j-t}\right)_{i,j\in\{1,\ldots,n\}} =t\sum_{k=0}^\infty \mathbf v_k\mathbf v_k^\top $$ is positive semidefinite. Now the result follows by passing the matrix on the LHS to the limit $t\to0$.
A big merit of the above proof is that it can be easily extended to prove the positive semidefiniteness of the power Cauchy matrix $\left(\frac{1}{(x_i+x_j)^p}\right)_{i,j\in\{1,\ldots,n\}}$ for any $p>0$.