At the very least, $\int u J_{2n}(u)\mathrm du$ for integer $n$ is expressible in terms of Bessel functions with some rational function factors.
To integrate $u J_0(u)$ for instance, start with the Maclaurin series:
$$u J_0(u)=u\sum_{k=0}^\infty \frac{(-u^2/4)^k}{(k!)^2}$$
and integrate termwise
$$\int u J_0(u)\mathrm du=\sum_{k=0}^\infty \frac1{(k!)^2}\int u(-u^2/4)^k\mathrm du$$
to get
$$\int u J_0(u)\mathrm du=\frac{u^2}{2}\sum_{k=0}^\infty \frac{(-u^2/4)^k}{k!(k+1)!}$$
thus resulting in the identity
$$\int u J_0(u)\mathrm du=u J_1(u)$$
For $\int u J_2(u)\mathrm du$, we exploit the recurrence relation
$$u J_2(u)=2 J_1(u)-u J_0(u)$$
and
$$\int J_1(u)\mathrm du=-J_0(u)$$
(which can be established through the series definition for Bessel functions) to obtain
$$\int u J_2(u)\mathrm du=-u J_1(u)-2J_0(u)$$
and in the general case of $\int u J_{2n}(u)\mathrm du$ for integer $n$, repeated use of the recursion relation
$$J_{n-1}(u)+J_{n+1}(u)=\frac{2n}{u}J_n(u)$$
as well as the additional integral identity
$$\int J_{2n+1}(u)\mathrm du=-J_0(u)-2\sum_{k=1}^n J_{2k}(u)$$
should give you expressions involving only Bessel functions.
On the other hand, $\int u J_{\nu}(u)\mathrm du$ for $\nu$ not an even integer cannot be entirely expressed in terms of Bessel functions; if $\nu$ is an odd integer, Struve functions are needed ($\int J_0(u)\mathrm du$ cannot be expressed solely in terms of Bessel functions, and this is where the Struve functions come in); for $\nu$ half an odd integer, Fresnel integrals are needed, and for general $\nu$, the hypergeometric function ${}_1 F_2\left({{}\atop b}{a \atop{}}{{}\atop c}\mid u\right)$ is required.
Using integral representation:
$$
J_0(x) = \frac{1}{2 \pi} \int_{-\pi}^\pi \mathrm{e}^{i x \sin \tau} \mathrm{d} \tau
$$
Thus the Fourier transform:
$$ \begin{eqnarray}
\mathcal{F}_x(J_0(x))(\omega) &=& \int_{-\infty}^\infty J_0(x) \mathrm{e}^{i \omega x} \mathrm{d} x \\ &=& \frac{1}{2 \pi} \int_{-\pi}^\pi \mathrm{d} \tau \, \mathcal{F}_x(\mathrm{e}^{i x \sin \tau})(\omega) \\
&=& \frac{1}{2\pi} \int_{-\pi}^\pi \mathrm{d} \tau \, \left( 2 \pi \right) \delta\left( \omega + \sin(\tau) \right) \\
&=& \int_{-\pi}^\pi \mathbf{1}_{-1 \le \omega \le 1} \delta\left( \omega + \sin(\tau) \right) \,\, \mathrm{d} \tau \\ &=& \int_{-\pi}^\pi \mathbf{1}_{-1 \le \omega \le 1} \frac{1}{\vert \cos(\tau) \vert} \left( \delta\left( \arcsin \omega + \tau \right) + \delta\left( \arcsin \omega - \operatorname{sign}(\omega) \pi + \tau \right) \right)\,\, \mathrm{d} \tau \\ &=& \mathbf{1}_{-1 \le \omega \le 1} \frac{2}{\sqrt{1-\omega^2}}
\end{eqnarray}
$$
Best Answer
Hint: Using the integral representation of the Bessel function $$J_n(x) = \frac{1}{2\pi} \int_{-\pi}^\pi dt\, e^{-i(n t - x \sin t)}$$ we find \begin{eqnarray*} \int_0^\infty dx\, e^{-p x} J_n(x) &=& \frac{1}{2\pi} \int_{-\pi}^\pi dt\, \frac{e^{-i n t}}{p-i\sin t}. \end{eqnarray*} This integral can be handled using the methods of residue calculus. Let $z=e^{-i t}$ so the contour is the unit circle. Only one of the poles lies inside the contour.
Addendum: The integral in terms of $z$ is $$\frac{1}{\pi i} \int_\Gamma dz\, \frac{z^n}{z^2+2p z-1},$$ where $\Gamma$ is the unit circle traversed in the counterclockwise sense. The poles are $$z_\pm = -p \pm \sqrt{1+p^2}.$$ Only $z_+$ lies inside the unit circle, so we pick up the residue at $z=z_+$.