Answer: (i) \(\frac{d}{dx}\{x(4+x^2)^{-n}\}=(4+x^2)^{-n}-2nx^2(4+x^2)^{-n-1}\), and \(8nI_{n+1}=(2n-1)I_n+2\times 8^{-n}\).

(ii) \(I_1=\frac{\pi}{8}\), and \(I_3=\frac{3\pi}{1024}+\frac{1}{128}\).

First differentiate using the product rule:

\(\frac{d}{dx}\{x(4+x^2)^{-n}\}=(4+x^2)^{-n}+x\{-n(4+x^2)^{-n-1}(2x)\}\).

Thus

\(\frac{d}{dx}\{x(4+x^2)^{-n}\}=(4+x^2)^{-n}-2nx^2(4+x^2)^{-n-1}\).

Now write \(x^2=(4+x^2)-4\). Then

\(-2nx^2(4+x^2)^{-n-1}=-2n\{(4+x^2)-4\}(4+x^2)^{-n-1}\).

So

\(\frac{d}{dx}\{x(4+x^2)^{-n}\}=(1-2n)(4+x^2)^{-n}+8n(4+x^2)^{-n-1}\).

Integrate both sides from \(0\) to \(2\):

\([x(4+x^2)^{-n}]_0^2=(1-2n)I_n+8nI_{n+1}\).

The left hand side is \(2(8)^{-n}\). Hence

\(2\times 8^{-n}=(1-2n)I_n+8nI_{n+1}\), so

\(8nI_{n+1}=(2n-1)I_n+2\times 8^{-n}\).

Next,

\(I_1=\int_0^2\frac{1}{4+x^2}\,dx\).

Using \(\int\frac{1}{x^2+a^2}\,dx=\frac1a\tan^{-1}\frac{x}{a}\), with \(a=2\),

\(I_1=\left[\frac12\tan^{-1}\frac{x}{2}\right]_0^2=\frac12\tan^{-1}1=\frac{\pi}{8}\).

Use the reduction formula with \(n=1\):

\(8I_2=I_1+2\times 8^{-1}=\frac{\pi}{8}+\frac14\).

Therefore

\(I_2=\frac{\pi}{64}+\frac{1}{32}\).

Now use the reduction formula with \(n=2\):

\(16I_3=3I_2+2\times 8^{-2}=3\left(\frac{\pi}{64}+\frac{1}{32}\right)+\frac{1}{32}\).

So

\(16I_3=\frac{3\pi}{64}+\frac18\), and hence

\(I_3=\frac{3\pi}{1024}+\frac{1}{128}\).