$\displaystyle \int_{0}^{x}J_\nu(t)dt=2\sum_{n=0}^{\infty}J_{\nu +2n+1}(x)$
Hint: Using the following recurrence relations show that both sides have the same derivative.
$\displaystyle J_{\nu-1}(x)+J_{\nu+1}(x)=\frac {2\nu}{x}J_\nu(x)$
$\displaystyle J_{\nu-1}(x)-J_{\nu+1}(x)=2J_\nu^{'}(x)$
You don't even need to use the first of the recurrence relations. Just notice that, according to the 2nd relation, \begin{align} \left(2\sum_{n=0}^{\infty}J_{\nu+2n+1}(x)\right)'&=\sum_{n=0}^{\infty}\Bigl(J_{\nu+2n}(x)-J_{\nu+2n+2}(x)\Bigr)=\\ &=J_{\nu}(x)-J_{\nu+2}(x)+J_{\nu+2}(x)-J_{\nu+4}(x)+J_{\nu+4}(x)-J_{\nu+6}(x)+\ldots=\\ &=J_{\nu}(x). \end{align}