Vtyjkyuk

How to get reduction formula of $$u_n=intfracx^nsqrtax^2+2bx+c$$

My try:

Here $P_n-1(x)$ is a polynomial of degree $(n-1)$
$$u_n=intfracx^nsqrtax^2+2bx+c=P_n-1(x)sqrtax^2+2bx+c+kintfracdxsqrtax^2+2bx+x$$
Differentiating:
$$x^n=P'_n-1(x)(ax^2+2bx+c)+frac12P_n-1(x)(2ax+2b)+k$$
$$implies k=0$$
I don't know how to proceed further:

Answer is of the form/ Spoiler:

$$(n+1)u_n+1+(2n+1)bu_n+ncu_n-1=?$$

You can Proceed in this way:

$$a, u_n+2=int fracax^n+2sqrtax^2+2bx+cdx$$

$$2b,u_n+1=int frac2bx^n+1sqrtax^2+2bx+cdx$$

$$c,u_n=int fraccx^nsqrtax^2+2bx+cdx$$

Adding all

$$a,u_n+2+2b,u_n+1+c,u_n=int x^nsqrtax^2+2bx+cdx=v_n tag1 $$

Using Integration by Parts to evaluate $$v_n=int x^nsqrtax^2+2bx+cdx=(sqrtax^2+2bx+c)fracx^n+1n+1-frac1n+1int fracx^n+1(ax+b) sqrtax^2+2bx+c$$

$implies$

$$v_n=(sqrtax^2+2bx+c)fracx^n+1n+1-fraca,u_n+2n+1-fracb,u_n+1n+1$$

Using above result of $v_n$ in $(1)$ we get

$$a,u_n+2+2b,u_n+1+c,u_n=(sqrtax^2+2bx+c)fracx^n+1n+1-fraca,u_n+2n+1-fracb,u_n+1n+1$$

$implies$

$$a(n+2)u_n+2+b(2n+3)u_n+1+c(n+1)u_n=x^n+1sqrtax^2+2bx+c$$