How to evaluate this integral containing a Bessel function: $\int_1^2 x^{-2} J_2(x)\,dx$? $$\int_1^2 x^{-2} J_2(x)\,dx$$
I can't solve this integration problem. I want to develop the expression using Derivation of the Bessel function and Recurrence relations between the Bessel functions and get the answer.

You can evaluate in terms of hypergeometric functions. Recall that
$$J_{\nu}(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{-z^2}{4}\right)$$
Therefore
$$x^{-2}J_2(x)=\frac{1}{x^2}\frac{(x/2)^2}{\Gamma(2+1)}~{}_0F_1\left(;2+1~;~\frac{-x^2}{4}\right) \\ =\frac{1}{8}~{}_0F_1\left(;3~;~\frac{-x^2}{4}\right)$$
Now, with the substitution $$-x^2/4=z\\  \mathrm dx=-\mathrm iz^{-1/2}~\mathrm dz$$ we get
$$\int x^{-2}J_2(x)\mathrm dx=\frac{-1}{8}\int {}_0F_1\left(;3~;~\frac{-x^2}{4}\right)\mathrm dx \\ =\frac{-\mathrm i}{8}\int z^{1/2-1}~{}_0F_1\left(;3~;~z\right)~\mathrm dz$$
Which from here we know has the primitive
$$=\frac{-\mathrm i}{8}\frac{z^{1/2}}{1/2}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~z\right)$$
Putting back our substitution $z=-x^2/4$ this is
$$=\frac{-\mathrm i}{8}(1/2)^{-1}\sqrt{\frac{-x^2}{4}}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right) \\ =\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)$$
I.E, that is
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-x^2}{4}\right)\right)=x^{-2}J_2(x)$$
Hence
$$\int_1^2 x^{-2}J_2(x)\mathrm dx=\frac{1}{4}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~-1\right)-\frac{1}{8}~{}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg |~\frac{-1}{4}\right)\\ \approx 0.102617$$
Confirmed by Mathematica:


Mathematica says
$${}_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2},3\end{matrix}~\bigg|~z\right)\\=\frac{2~I_0(2\sqrt z)\cdot\big(-1+4z+2\pi z~L_1(2\sqrt z)\big)}{3z}-\frac{2~I_1(2\sqrt z)\cdot\big(-1+2z+2\pi z^{3/2}L_0(2\sqrt z)\big)}{3z^{3/2}}$$
Where
$$I_\nu(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{z^2}{4}\right)$$
Is a Modified Bessel function
and
$$L_\nu(z)=\frac{z~(z/2)^\nu}{\sqrt{\pi}~\Gamma(\nu+3/2)}~{}_1F_2\left(\begin{matrix}1 \\ \frac{3}{2},\nu+\frac{3}{2}\end{matrix}~\bigg|~\frac{z^2}{4}\right)$$
Is a modified Struve function.
But IMO this doesn't qualify as "simpler".
