Cfrgtkky

So I tried proving this integral reduction formula but to no avail:

If $$I_n=int frac{x^n}{sqrt{ax+b}}dx$$, then
$$int frac{x^n}{sqrt{ax+b}}dx=frac{2x^nsqrt{ax+b}}{a(2n+1)}-frac{2nb}{a(2n+1)}I_{n-1}$$
I tried integrating by parts and my attempt went as followed:
$$int frac{x^n}{sqrt{ax+b}}dxbegin{vmatrix}u=x^n\du=nx^{n-1}dxend{vmatrix}v=int frac{1}{sqrt{ax+b}}dxenspace v=frac{2sqrt{ax+b}}{a}\
int udv=uv-int vdu$$
The integral then becomes
$$int frac{x^n}{sqrt{ax+b}}dx=frac{2x^nsqrt{ax+b}}{a}-frac{2n}{a}int x^{n-1}sqrt{ax+b}dx$$
Multiplying and dividing the integrand on the right by $sqrt{ax+b}$ gives
$$int frac{x^n}{sqrt{ax+b}}dx=frac{2x^nsqrt{ax+b}}{a}-frac{2n}{a}int frac{x^{n-1}(ax+b)}{sqrt{ax+b}}dx\
=frac{2x^nsqrt{ax+b}}{a}-2nint frac{x^n}{sqrt{ax+b}}dx-frac{2nb}{a}int frac{x^{n-1}}{sqrt{ax+b}}dx\
=frac{2x^nsqrt{ax+b}}{a}-2nI_n-frac{2nb}{a}I_{n-1}$$
So as you can see, I tried algebraic manipulation in order to yield an $I_{n-1}$ term within the integral, but this was as far as I got and I don't know how to proceed from here (or if my approach was even correct). Can I get some help with this one?

You are almost there. Just move the $-2nI_n$ term to the other side, then divide everything by $2n+1$

$$I_n=frac{2x^nsqrt{ax+b}}{a}-2nI_n-frac{2nb}{a}I_{n-1}\
I_n+2nI_n=frac{2x^nsqrt{ax+b}}{a}-frac{2nb}{a}I_{n-1}\
I_n=frac{2x^nsqrt{ax+b}}{a(2n+1)}-frac{2nb}{a(2n+1)}I_{n-1}$$