Reduction Integration Question ($\int_0^1{\frac{x^n}{\sqrt{x+1}} dx}$)

Question

What integration by parts things do I use for a reduction formula for $\int_0^1{\frac{x^n}{\sqrt{x+1}} dx}$?

I have tried many different ways of obtaining it without success.

Answer 1

$$I_n=\int\frac{x^ndx}{\sqrt{x+1}}=\int\frac{x^{n-1}(1+x-1)dx}{\sqrt{x+1}} $$

$$=\int x^{n-1}\sqrt{x+1}dx-\int\frac{x^{n-1}dx}{\sqrt{x+1}}=\int x^{n-1}\sqrt{x+1}dx-I_{n-1}$$

Again, $$\int x^{n-1}\sqrt{x+1}dx= \sqrt{x+1}\frac{x^n}n-\int \frac{x^n}{2n\sqrt{x+1}}dx=\sqrt{x+1}\frac{x^n}n-\frac{I_n}{2n}$$

$$\implies I_n=\sqrt{x+1}\frac{x^n}n-\frac{I_n}{2n}-I_{n-1}$$

Answer 2

Just another way of reduction

\begin{align} I_n=&\int_{0}^{1}\dfrac{x^n}{\sqrt{1+x}}dx\\ \ =&2\int_{0}^{\pi/4}\tan^{2n+1}\theta \sec \theta \ d\theta\\ \ =&2\int_{0}^{\pi/4}\tan^{2n}\theta\cdot \tan \theta\sec \theta \ d\theta\\ \ =&2\big|\tan^{2n}\theta \sec \theta\big|_{0}^{\pi/4}-4n\int_{0}^{\pi/4}\tan^{2n-1}\theta \sec^3\theta \ d\theta\\ \ =& 2\sqrt{2}-2n\left(I_{n-1}+I_n\right)\\ \ =& \frac{2\sqrt{2}}{2n+1}-\frac{2n}{2n+1}I_{n-1} \end{align} Now carry on from here to get $I_n$.