<s>[INST] Difficulty proving integral reduction formula So I tried proving this integral reduction formula but to no avail:
If $$I_n=\int \frac{x^n}{\sqrt{ax+b}}dx$$, then 
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$
I tried integrating by parts and my attempt went as followed:
$$\int \frac{x^n}{\sqrt{ax+b}}dx\begin{vmatrix}u=x^n\\du=nx^{n-1}dx\end{vmatrix}v=\int \frac{1}{\sqrt{ax+b}}dx\enspace v=\frac{2\sqrt{ax+b}}{a}\\
\int udv=uv-\int vdu$$
The integral then becomes
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\sqrt{ax+b}dx$$
Multiplying and dividing the integrand on the right by $\sqrt{ax+b}$ gives
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int \frac{x^{n-1}(ax+b)}{\sqrt{ax+b}}dx\\
=\frac{2x^n\sqrt{ax+b}}{a}-2n\int \frac{x^n}{\sqrt{ax+b}}dx-\frac{2nb}{a}\int \frac{x^{n-1}}{\sqrt{ax+b}}dx\\
=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}$$ 
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?
 [/INST] You are almost there. Just move the $-2nI_n$ term to the other side, then divide everything by $2n+1$ 
$$I_n=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}\\
I_n+2nI_n=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2nb}{a}I_{n-1}\\
I_n=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$ 
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