= Solution
Let
$$
u(x)=J_0(\alpha x),
\qquad
v(x)=J_0(\beta x).
$$
Their <Bessel differential equation>[Bessel equations] in self-adjoint form are
$$
(xu')'+\alpha^2xu=0,
\qquad
(xv')'+\beta^2xv=0.
$$
Multiply the first by $v$, the second by $u$, subtract, and integrate. The left side becomes a boundary term:
$$
\left[x(vu'-uv')\right]_0^1
+(\alpha^2-\beta^2)\int_0^1uvx\,dx=0.
$$
Regularity at zero makes the lower boundary term vanish, while
$$
u'(1)=\alpha J_0'(\alpha),
\qquad
v'(1)=\beta J_0'(\beta).
$$
Therefore
$$
\boxed{
\int_0^1J_0(\alpha x)J_0(\beta x)x\,dx
=\frac{\beta J_0(\alpha)J_0'(\beta)
-\alpha J_0(\beta)J_0'(\alpha)}
{\alpha^2-\beta^2}}.
$$
If $\alpha=\gamma_k$ and $\beta=\gamma_\ell$ with $k\ne\ell$, both endpoint values of $J_0$ vanish, so the integral is zero. This is the weighted <orthogonality> of distinct eigenfunctions in a <Sturm-Liouville problem>. For the norm, hold $\alpha=\gamma_k$ and let $\beta\to\gamma_k$. Since
$$
J_0(\beta)=J_0'(\gamma_k)(\beta-\gamma_k)
+o(\beta-\gamma_k),
$$
the quotient tends to
$$
\boxed{\int_0^1J_0(\gamma_kx)^2x\,dx
=\frac12J_0'(\gamma_k)^2}.
$$
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