$By\ recurrence\ formuls\ \left(VI\right),\ we\ have,{\large } \\[2ex] \displaystyle \frac{d}{dx}\left[-x^{-n}J_n\left(x\right)\right]=-x^{-n}J_{n+1}(x) \\[2ex] $

$\displaystyle Integrating\ both\ sides,\ we\ get, \\[2ex] \displaystyle \int-x^{-n}J_{n+1}\left(x\right)dx=\ -x^{-n}J_n\left(x\right) \\[2ex] $

$\displaystyle Now,\ write\ the\ lhs\ of\ the\ required\ result\ as \\[2ex] \displaystyle \int J_3\left(x\right)dx=\ \int x^2[x^{-2}J_3\left(x\right)]dx \\[2ex] \displaystyle Integrating\ the\ rhs\ by\ parts, \\[2ex] \displaystyle \int J_3\left(x\right)dx=\ x^2\int x^{-2}J_3\left(x\right)dx-\int \left[\int x^{-2}J_3\left(x\right)dx(2x)\right]dx \\[2ex] {\large } $

$\displaystyle Putting\ n=2\ in\ \left(1\right)above, \\[2ex] \displaystyle \int x^{-2}J_3\left(x\right)dx={-x}^{-2}J_2\left(x\right) \\[2ex] {\large \displaystyle \therefore{}\int J_3\left(x\right)dx\\[3ex] \displaystyle =x^2\left(-x^{-2}\right)J_2\left(x\right)-\int-x^{-2}J_2\left(x\right)(2x)dx}\\[2ex] \displaystyle =-J_2\left(x\right)+2\int\int x^{-1}J_2\left(x\right)dx \\[2ex] $

$\displaystyle Putting\ n=1\ in\ \left(1\right)above, \\[2ex] \displaystyle \int x^{-1}J_2\left(x\right)dx={-x}^{-1}J_1\left(x\right) \\[2ex] \displaystyle \therefore{}\int J_3\left(x\right)dx=\ -\frac{2J_1(x)}{x}-J_2(x) \\[2ex]$