Prove that $\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2}J_2(x) +c$ 
Prove that  $$\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac
 {8}{x^2}J_2(x) +c$$ where $J_n (x)$ is the Bessel function of first kind
  and order $n $.

My attempt:
I know the following recurrence relations:
$$\frac {d}{dx}\left[x^nJ_n (x)\right]=x^nJ_{n-1} (x)$$
$$\frac {d}{dx}\left[x^{-n}J_n (x)\right]=-x^{-n}J_{n+1} (x)$$
$$2J'_n(x)=J_{n-1} (x)-J_{n+1} (x)$$
$$2\frac {n}{x}J_n(x)=J_{n-1} (x)+J_{n+1} (x)$$
But I cannot figure out how to get the above expression. Please help.
 A: I will use the second of your recurrence relations:
$$
\frac{d}{dx}[x^{-n}J_n(x)] = -x^{-n}J_{n + 1}(x)
$$
And consider the next cases


*

*$n = 2$
$$
 \frac{d}{dx}[x^{-2}J_2(x)] = -x^{-2}J_{3}(x) ~~~\Rightarrow~~~ \int dx\; x^{-2}J_3(x) = -x^{-2}J_2(x) \tag{1}
 $$

*$n = 3$
$$
 \frac{d}{dx}[x^{-3}J_3(x)] = -x^{-3}J_{4}(x) ~~~\Rightarrow~~~ -x^{2}\frac{d}{dx}[x^{-3}J_3(x)] = x^{-1}J_{4}(x)
 $$
Therefore
\begin{eqnarray}
 \int dx\; x^{-1}J_{4}(x) &=& -\int dx\; x^{2}\frac{d}{dx}[x^{-3}J_3(x)] ~~~\mbox{make}~~~ u = x^2\mbox{ and } dv = d(x^{-3}J_3(x)) \\
 &=& -x^{-1}J_3(x) + \int dx\; 2x^{-2}J_3(x) \\
 &\stackrel{(1)}{=}& -x^{-1}J_3(x) + 2[-x^{-2}J_2(x)] \tag{2}
 \end{eqnarray}

*$n = 4$
$$
 \frac{d}{dx}[x^{-4}J_4(x)] = -x^{-4}J_{5}(x) ~~~\Rightarrow~~~ -x^4\frac{d}{dx}[x^{-4}J_4(x)] = J_{5}(x)
 $$
or equivalently
\begin{eqnarray}
 \int dx\;J_5(x) &=& -\int dx\; x^4\frac{d}{dx}[x^{-4}J_4(x)] ~~~\mbox{call}~~~ u = x^4\mbox{ and } dv = d(x^{-4}J_4(x)) \\
&=& - J_4(x) + \int dx\; 4 x^{-1}J_4(x) \\
&\stackrel{(2)}{=}&  -J_4(x) + 4\left[-x^{-1}J_3(x) -2 x^{-2}J_2(x) \right] \\
&=& -J_4(x) -\frac{4}{x}J_3(x) - \frac{8}{x^2}J_2(x)
 \end{eqnarray}
Add the integration constant to get
$$
\int dx\; J_5(x) = -J_4(x) -\frac{4}{x}J_3(x) - \frac{8}{x^2}J_2(x) + c
$$
A: Following Alex R's comment, I am proceeding as follows:
$$\int J_5(x) dx$$
$$=\int \frac {x^4}{x^4} J_5(x) dx$$
$$=\int x^4 \cdot x^{-4} J_5(x) dx$$
Integrating by parts, we get
$$x^4 \cdot \int x^{-4} J_5(x) dx - \int \left\{\frac {d}{dx}(x^4) \cdot \int x^{-4} J_5(x) dx \right\} dx$$
$$=x^4 \cdot  (-1) \cdot x^{-4} J_4(x) - \int \left\{4x^3 \cdot (-1) \cdot x^{-4} J_4(x)\right\} dx$$
$$=-J_4(x) +4 \int \left\{x^3 \cdot x^{-4} J_4(x)\right\} dx$$
$$=-J_4(x) +4 \int \left\{x^2 \cdot x^{-3} J_4(x)\right\} dx$$
Similarly,  again integrating by parts, we get
$$=-J_4(x) +4 x^2 \cdot \int x^{-3} J_4(x) dx - 4\cdot \int \left\{\frac {d}{dx}(x^2) \cdot \int x^{-3} J_4(x) dx\right\} dx$$
$$=-J_4(x) +4 x^2 \cdot (-1) \cdot x^{-3} J_3(x) - 4\cdot \int \left\{2x \cdot (-1) \cdot x^{-3} J_3(x)\right\} dx$$
$$=-J_4(x) - \frac {4}{x} J_3(x) + 8\cdot \int \left\{x^{-2} \cdot  J_3(x)\right\} dx$$
$$=-J_4(x) - \frac {4}{x} J_3(x) - 8\cdot \int \left\{x^{-2} \cdot  J_2(x)\right\}$$
$$=-J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2} J_2(x) +c$$
Hence the relation is proved.
Thanks Alex R for the great hint.
