Proof:

Considering the l.h.s of the given equation

$\int\;x^4J_1(x)=\:\int\;x^2\times x^2J_1(x)$

$considering \;x^2=u\;and\;x^2J_1(x)=v\;and\;applying\;u\times v\;rule$

$\int\;x^4J_1(x).dx=x^2\int\;x^2J_1(x).dx\;-{\int\;[\dfrac {d} {dx}x^2\times { \int\;x^2J_1(x)dx}]\;dx}$

$\int\;x^4J_1(x)=\;x^2\times x^2J_2(x)-\int\;2x\times x^2J_2(x)dx$

$the \;above \;equation \;is\; by \;formula \;(\int\;x^nJ_{n-1}=x^nJ_n(x))$

$\int\;x^4J_1(x)=x^4J_2(x)-\int\;2x^3J_2(x)dx$

$using \;the \;formula \;again\;we\;get$

$\int\;x^4J_1(x)=\;x^4J_2(x)-2x^3J_3(x)+c$

hence proved