Answer: (a) \(I_{-1}=\ln 3\).

(b) \((n+1)I_n=nI_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^n\).

(c) \(s=\frac{1}{2}I_1\), and hence \(s=\frac{1}{4}\ln 3+\frac{5}{9}\).

(a) For \(n=-1\),

\(I_{-1}=\int_0^{4/3}(1+x^2)^{-1/2}\,dx.\)

Using \(\int \frac{1}{\sqrt{1+x^2}}\,dx=\sinh^{-1}x=\ln\left(x+\sqrt{1+x^2}\right)\),

\(I_{-1}=\left[\sinh^{-1}x\right]_0^{4/3}=\ln\left(\frac{4}{3}+\sqrt{1+\frac{16}{9}}\right)-\ln(1).\)

Since \(\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac{5}{3}\),

\(I_{-1}=\ln\left(\frac{4}{3}+\frac{5}{3}\right)=\ln 3.\)

(b) Consider \(x(1+x^2)^{n/2}\). Differentiating,

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

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

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

So

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

Integrating from \(0\) to \(\frac{4}{3}\),

\(\left[x(1+x^2)^{n/2}\right]_0^{4/3}=(n+1)I_n-nI_{n-2}.\)

Evaluating the left-hand side,

\(\left[x(1+x^2)^{n/2}\right]_0^{4/3}=\frac{4}{3}\left(1+\frac{16}{9}\right)^{n/2}=\frac{4}{3}\left(\frac{25}{9}\right)^{n/2}=\frac{4}{3}\left(\frac{5}{3}\right)^n.\)

Hence

\((n+1)I_n-nI_{n-2}=\frac{4}{3}\left(\frac{5}{3}\right)^n,\)

so

\((n+1)I_n=nI_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^n.\)

(c) For \(y=x^2\), \(\frac{dy}{dx}=2x\), so the arc length is

\(s=\int_0^{2/3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\int_0^{2/3}\sqrt{1+4x^2}\,dx.\)

Using the substitution \(u=2x\), we have \(dx=\frac{1}{2}du\). Also, when \(x=0\), \(u=0\), and when \(x=\frac{2}{3}\), \(u=\frac{4}{3}\). Therefore

\(s=\frac{1}{2}\int_0^{4/3}\sqrt{1+u^2}\,du=\frac{1}{2}I_1.\)

Now use the relation from part (b) with \(n=1\):

\(2I_1=I_{-1}+\frac{4}{3}\left(\frac{5}{3}\right)=\ln 3+\frac{20}{9}.\)

So

\(I_1=\frac{1}{2}\ln 3+\frac{10}{9}.\)

Hence

\(s=\frac{1}{2}I_1=\frac{1}{4}\ln 3+\frac{5}{9}.\)