Answer: By differentiating \(\dfrac{x}{(1+x^4)^n}=x(1+x^4)^{-n}\), we get

\(\dfrac{\mathrm d}{\mathrm d x}\bigl(x(1+x^4)^{-n}\bigr)=(1+x^4)^{-n}-4nx^4(1+x^4)^{-n-1}.\)

Rewriting the first term gives

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

Integrating from \(0\) to \(1\),

\(\left[\dfrac{x}{(1+x^4)^n}\right]_0^1=(1-4n)I_n+4nI_{n+1}.\)

The left-hand side is

\(\dfrac{1}{2^n}-0=2^{-n},\)

so

\(4nI_{n+1}=2^{-n}+(4n-1)I_n.\)

Using \(I_1=0.86697\):

For \(n=1\), \(4I_2=\dfrac12+3I_1\), hence \(I_2=\dfrac{\frac12+3(0.86697)}{4}=0.7752275.\)

For \(n=2\), \(8I_3=\dfrac14+7I_2\), hence \(I_3=\dfrac{\frac14+7(0.7752275)}{8}=0.7095740625.\)

Therefore \(I_3=0.70957\) to 5 decimal places, so \(I_3\approx 0.7096\).

Let \(f(x)=x(1+x^4)^{-n}\). Then

\(f'(x)=(1+x^4)^{-n}+x\cdot(-n)(1+x^4)^{-n-1}\cdot 4x^3\)

\(\phantom{f'(x)}=(1+x^4)^{-n}-4nx^4(1+x^4)^{-n-1}.\)

Now write \(x^4=(1+x^4)-1\) in the second term:

\(f'(x)=(1+x^4)^{-n}-4n\bigl((1+x^4)-1\bigr)(1+x^4)^{-n-1}\)

\(\phantom{f'(x)}=(1-4n)(1+x^4)^{-n}+4n(1+x^4)^{-n-1}.\)

Integrate from \(0\) to \(1\):

\(\int_0^1 f'(x)\,\mathrm dx=\left[\dfrac{x}{(1+x^4)^n}\right]_0^1.\)

So

\(\left[\dfrac{x}{(1+x^4)^n}\right]_0^1=(1-4n)\int_0^1\dfrac{1}{(1+x^4)^n}\,\mathrm dx+4n\int_0^1\dfrac{1}{(1+x^4)^{n+1}}\,\mathrm dx.\)

Using the definition of \(I_n\), this becomes

\(\left[\dfrac{x}{(1+x^4)^n}\right]_0^1=(1-4n)I_n+4nI_{n+1}.\)

At \(x=1\), \(\dfrac{x}{(1+x^4)^n}=\dfrac{1}{2^n}\), and at \(x=0\) it is \(0\). Hence

\(2^{-n}=(1-4n)I_n+4nI_{n+1}.\)

Rearranging gives

\(4nI_{n+1}=2^{-n}+(4n-1)I_n,\)

as required.

Now use the recurrence to find \(I_3\).

For \(n=1\): \(4I_2=\dfrac12+3I_1\), so

\(I_2=\dfrac{\frac12+3(0.86697)}{4}=0.7752275.\)

For \(n=2\): \(8I_3=\dfrac14+7I_2\), so

\(I_3=\dfrac{\frac14+7(0.7752275)}{8}=0.7095740625.\)

Therefore \(I_3=0.70957\) correct to 5 decimal places.