We know that $\frac{d}{dx}x^{-n} J_{n}(x) = -x^{n}J_{n + 1}(x)$

Integrating both sides we get,

$\int \frac{d}{dx}x^{-n} J_{n}(x) = \int -x^{-n}J_{n + 1}(x)\\ \therefore -x^{-n}J_{n}(x) = \int -x^{-n}J_{n + 1}(x)dx......(A)$

Now L.H.S of the required result,

i.e.

$\int J_{3}(x)dx = \int x^2 (x^{-2}J_{3}(x))dx$

Now integrating the R.H.S. by part

$\therefore \int J_{3}(x)dx = x^2 \int x^{-2} J_{3}(x) dx - 2 \int \lfloor x \int x^{-2} J_{3}(x)dx\rfloor$

[using Leibnitz rule].......(B)

Now put n = 2 in equation (A)

$\therefore -x^{-2}J_{2}(x) = \int x^{-2}J_{3}(x)dx$

Substituting the above value in equation (B)

$\therefore J_{3}(x) dx = x^2 (-x^{-2}J_{2}(x)) - 2x \int -x^{-2} J_{2}(x) dx\\ J_{3}(x)dx = - J_{2}(x) - 2 \int x^{-1} J_{2}(x) dx........(C)$

Now put n = 1 in equation (A)

$\therefore -x^{-1}J_{1}(x) = \int x^{-1}J_{2}(x)dx$

Substituting the above value in equation (C)

$\therefore J_{3}(x)dx = -J_{2}(x) - 2[-x^{-1} J_{1}(x)] \\ \int J_{3}(x)dx = \frac{2}{x} J_{1}(x) - J_{2}(x)....(D)$

Now in order to get the result as express the result in terms of $J_{0}$ and $J_{1}$ we need to express $J_{2}(x)$ in terms of $J_{0}$ and $J_{1}$.

Hence, we know that

$2nJ_{n}(x) = xJ_{n + 1}(x) + xJ_{n - 1}(x)$

[recurrence formula (4)]

$\therefore J_{n + 1}(x) = \frac{2x}{n}J_{n}(x) - J_{n - 1}(x)$

Put n = 2

$\therefore J_{2}(x) = \frac{2}{x}J_{1}(x) - J_{0}(x)............(E)$

Substituting the value $J_{2}(x) from equation (E) in equation (D) we get, $\int J_{3}(x)dx = \frac{-2}{x}J_{1}(x) - \bigg(\frac{2}{x} J_{1}(x) - J_{0}(x)\bigg)\ \int J_{3}(x)dx = \frac{-4}{x}J_{1}(x) + J_{0}(x)$ The above expression represents $J_{0}(x)dx$ in terms of $J_{0}(x)$ and $J_{1}(x)$ as required.