Deriving a reduction formula for the integral: $I_n=\int _0^1 \left(1+x^2\right)^n \mathrm{d}x$

Question

I'm trying to derive a reduction formula for the following integral:

$$I_n=\int _0^1 \left(1+x^2\right)^n \mathrm{d}x$$

So far, I have tried applying integration by parts and have reached till:

$$I_n= x(1+x^2)^n\bigg|_0^1-2n\int_0^1 x^2\cdot (1+x^2)^{n-1} \mathrm{d}x$$

I am not able to proceed after this. The integral could be written in terms of $I_{n-1}$ but with the $x^2$ term with it, I'm not really sure how to proceed. I have also tried substituting $x=\tan t$ but I'm not able to derive anything significant from this either.

I would like some hints in which direction I should proceed either in the first method or in the method I have substituted for $\tan t$.
Thanks in advance.

Answer 1

Continuing the derived integral: $$I_n=(1+x^2)^n\bigg|_0^1-2n\int_0^1x^2\cdot (1+x^2)^{n-1}\mathrm{d}x$$ $$\Rightarrow I_n=2^n-2n\int_0^1(x^2+1-1)\cdot (1+x^2)^{n-1}\mathrm{d}x$$ $$\Rightarrow I_n=2^n-2n\left[\int_0^1(1+x^2)^n\mathrm{d}x-\int_0^1(1+x^2)^{n-1}\mathrm{d}x\right]$$ $$\Rightarrow I_n=2^n-2n\left[I_n-I_{n-1}\right]$$ $$\Rightarrow I_n =2^n-2nI_n+2nI_{n-1}$$ Solving the above linear equation in $I_n$ gives us : $$\boxed{I_n(2n+1)-2nI_{n-1}=2^n}$$

Answer 2

I would like to investigate their difference which help us evaluate the integral easily. $$ \begin{aligned} I_{n}-I_{n-1} &=\int_{0}^{1} x^{2}\left(1+x^{2}\right)^{n-1} d x \\ & \stackrel{IBP}{=} \frac{1}{2 n} \int_{0}^{1} x d\left(1+x^{2}\right)^{n} \\ &=\left[\frac{x\left(1+x^{2}\right)^{n}}{2 n}\right]_{0}^{1}-\frac{1}{2 n} \int_{0}^{1}\left(1+x^{2}\right)^{n} d x \\ &=\frac{2^{n-1}}{n}-\frac{1}{2 n} I_{n} \end{aligned} $$

Rearranging yields the reduction formula $$ \boxed{I_{n}= \frac{1}{2 n+1}\left(2^{n}+2 n I_{n-1}\right)} $$