We have

$\int\ x^n\ J_{n-1} (x)\ dx=x^n\ J_n(x)$

x       0         b

t        0        ab 

Consider $I=\int_0^{b}x\ J_0\ (ax)dx$

Put

ax = t

a dx = dt

$dx=\dfrac{dt}{a}$

we get, 

$I=\int_0^{ab}\dfrac{t}{a}.J_0(t)\dfrac{dt}{a}$

$I=\dfrac{1}{a^2}[t.J_1(t)]_0^{ab}dt=\dfrac{1}{a^2}[abJ_1(ab)-0]$

$I=\dfrac{b}{a}J_1(ab)$

$\int_0^{ab}x\ J_0(ax)dx=\dfrac{b}{a}J_1(ab)$