How can I find a reduction formula for the following

Question

How can I find the reduction formula for the integral below?

$${I}_n=\int \frac{x^{n}}{\sqrt{2x+1}}dx $$

I have literally no idea how to solve this question and would appreciate any pointers in the right direction.

Answer 1

Do integration by parts with $u=x^n$ and $v'=(2x+1)^{-\frac 12}$

So you get $$I_n=x^n\sqrt{2x+1}-n\int x^{n-1}\sqrt{2x+1}dx$$

Then write $$\sqrt{2x+1}=\frac{2x+1}{\sqrt{2x+1}}$$ and split up the integral into two pieces, which will give $$I_n=x^n\sqrt{2x+1}-2nI_n-nI_{n-1}$$

And there you are...

Answer 2

$$\int \frac{x^n}{\sqrt{2x+1}}\,dx \stackrel{x\mapsto\frac{y}{2}}{=} \frac{1}{2^{n+1}}\int \frac{y^n}{\sqrt{y+1}}\,dy \stackrel{y\mapsto z-1}{=} \frac{1}{2^{n+1}}\int \frac{(z-1)^n}{\sqrt{z}}\,dz $$ and by the binomial theorem the last integral equals $$ \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k} \int z^{k-1/2}\,dz = \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}\frac{z^{k+1/2}}{k+1/2}.$$