Q:

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:

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.