Answer: Using integration by parts on \(I_n=\int_0^1 \frac{1}{(1+x^2)^n}\,dx\), take \(u=(1+x^2)^{-n}\) and \(dv=dx\). Then \(du=-2nx(1+x^2)^{-(n+1)}dx\) and \(v=x\).

So

\(I_n=\left[\frac{x}{(1+x^2)^n}\right]_0^1+2n\int_0^1 \frac{x^2}{(1+x^2)^{n+1}}\,dx.\)

Now \(\left[\frac{x}{(1+x^2)^n}\right]_0^1=2^{-n}\), and since \(x^2=(1+x^2)-1\),

\(\int_0^1 \frac{x^2}{(1+x^2)^{n+1}}\,dx=\int_0^1 \left(\frac{1}{(1+x^2)^n}-\frac{1}{(1+x^2)^{n+1}}\right)dx=I_n-I_{n+1}.\)

Hence

\(I_n=2^{-n}+2n(I_n-I_{n+1}).\)

Rearranging gives

\(2nI_{n+1}=2^{-n}+(2n-1)I_n.\)

Given \(I_1=\frac{\pi}{4}\), with \(n=1\):

\(2I_2=\frac12+\frac{\pi}{4}\), so \(I_2=\frac14+\frac{\pi}{8}.\)

With \(n=2\):

\(4I_3=\frac14+3I_2=\frac14+3\left(\frac14+\frac{\pi}{8}\right)=1+\frac{3\pi}{8}.\)

Therefore

\(I_3=\frac14+\frac{3\pi}{32}.\)

Let \(I_n=\int_0^1 \frac{1}{(1+x^2)^n}\,dx\). We first derive the reduction formula.

Use integration by parts with \(u=(1+x^2)^{-n}\) and \(dv=dx\). Then \(v=x\) and \(du=-2nx(1+x^2)^{-(n+1)}dx\). Hence

\(I_n=\int_0^1 (1+x^2)^{-n}dx=\left[\frac{x}{(1+x^2)^n}\right]_0^1+2n\int_0^1 x^2(1+x^2)^{-(n+1)}dx.\)

The boundary term is

\(\left[\frac{x}{(1+x^2)^n}\right]_0^1=\frac{1}{2^n}.\)

Now write \(x^2=(1+x^2)-1\). Then

\(x^2(1+x^2)^{-(n+1)}=(1+x^2)^{-n}-(1+x^2)^{-(n+1)}.\)

So

\(\int_0^1 x^2(1+x^2)^{-(n+1)}dx=\int_0^1 \frac{1}{(1+x^2)^n}dx-\int_0^1 \frac{1}{(1+x^2)^{n+1}}dx=I_n-I_{n+1}.\)

Substituting back,

\(I_n=2^{-n}+2n(I_n-I_{n+1}).\)

Rearrange:

\(I_n=2^{-n}+2nI_n-2nI_{n+1}\)

so

\(2nI_{n+1}=2^{-n}+(2n-1)I_n.\)

This proves the required formula.

Now use \(I_1=\frac{\pi}{4}\) to find \(I_3\).

For \(n=1\):

\(2I_2=2^{-1}+(2\cdot1-1)I_1=\frac12+\frac{\pi}{4}.\)

Therefore

\(I_2=\frac14+\frac{\pi}{8}.\)

For \(n=2\):

\(4I_3=2^{-2}+(2\cdot2-1)I_2=\frac14+3\left(\frac14+\frac{\pi}{8}\right).\)

Simplifying,

\(4I_3=\frac14+\frac34+\frac{3\pi}{8}=1+\frac{3\pi}{8}.\)

Hence

\(I_3=\frac14+\frac{3\pi}{32}.\)