Determinant (or positive definiteness) of a Hankel matrix

Question

I need to prove that the Hankel matrix given by $a_{ij}=\frac{1}{i+j}$ is positive definite. It turns out that it is a special case of the Cauchy matrices, and the determinant is given by the Cauchy determinant. But I also want to use some problem-specific techniques.
Inspired by the Hilbert matrix, I want to find a function $u(x)$, such that $\int_0^1 x^{k}u(x)dx=1/k$ for any $k \in \mathbb{N}$, so that $a_{ij}=\int_0^1 x^{i}x^{j}u(x)dx$, and $A$ can be considered as the Gramian of $\{ 1, x, \dots, x^n \}$ in the space of polynomials, with the inner product $(f,g)=\int_0^1 f(x)g(x)u(x)dx$. Is it possible?

Answer 1

As user "daw" points out in the comment section, you may consider $a_{ij} = \frac1{i+j} = \int_0^1 t^{i+j-1} dt$. That $A$ is positive semidefinite is trivial, but you need to argue why it is invertible (hence positive definite). Properties of Vandermonde matrices would be useful.