Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx = \frac 14 + \frac {\pi}8$

Question

Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx=\frac 14 + \frac {\pi}8$

My attempt (integration by parts):

\begin{align} I_n & = \int_0^1 \frac 1{(1+x^2)^n}\,dx = \left. \frac {x}{(1+x^2)^n} \right|_0^1+2n\int_0^1 \frac {x^2+1-1}{(1+x^2)^{n+1}}\,dx \\[10pt] & =\frac 1{2^n}+2n \times I_n-2n\times I_{n+1}\implies I_{n+1}= \frac 1{2^{n+1}n}+\frac {2n-1}{2n}I_n. \end{align}

where $I_1=\frac {\pi}{4}.$

Finding out that $I_2=\frac {\pi}{8}+\frac 14.$ But how do I find out that this is the only solution?

Answer 1

Hint. The sequence $I_n$ is strictly decreasing because for $x\in (0,1]$, $$0<\frac {1}{1+x^2} <1.$$ .

Answer 2

Note that the sequence $(I_n)$ is decreasing, hence there is at most one value of $n$ such that $I_n=\frac{1}{4}+\frac{\pi}{8}$.