Answer: (a) Using integration by parts with \(u=x^n\) and \(dv=(1+2x)^{-\frac12}\,dx\), we can derive \((2n+1)I_n=\sqrt{3}-nI_{n-1}\) for \(n\ge 1\).

(b) Hence \(I_0=\sqrt{3}-1\), so \(I_1=\frac13\), \(I_2=\frac{\sqrt3}{5}-\frac{2}{15}\), and therefore \(I_3=\frac{2}{35}(\sqrt3+1)\).

Let

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

To obtain a recurrence, write the integrand in a form suitable for integration by parts:

\(I_n=\int_0^1 x^n(1+2x)^{-\frac12}\,dx\).

Take \(u=x^n\) and \(dv=(1+2x)^{-\frac12}\,dx\). Then \(du=nx^{n-1}dx\), and since

\(\int (1+2x)^{-\frac12}dx=(1+2x)^{\frac12}\),

we may take \(v=(1+2x)^{\frac12}\).

So

\(I_n=\left[x^n(1+2x)^{\frac12}\right]_0^1-\int_0^1 n x^{n-1}(1+2x)^{\frac12}\,dx\).

Now rewrite the integral in the second term by splitting the factor \(1+2x\):

\(\int_0^1 n x^{n-1}(1+2x)^{\frac12}\,dx = n\int_0^1 \frac{x^{n-1}(1+2x)}{\sqrt{1+2x}}\,dx\)

\(=n\int_0^1 \frac{x^{n-1}}{\sqrt{1+2x}}\,dx+2n\int_0^1 \frac{x^n}{\sqrt{1+2x}}\,dx\)

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

Also,

\(\left[x^n(1+2x)^{\frac12}\right]_0^1=\sqrt3\),

since the lower limit is \(0\) and the upper limit gives \(1\cdot\sqrt3\).

Therefore

\(I_n=\sqrt3-nI_{n-1}-2nI_n\).

Rearranging gives

\((2n+1)I_n=\sqrt3-nI_{n-1}\),

as required.

Now find \(I_3\) by working up from \(I_0\).

First,

\(I_0=\int_0^1 \frac{1}{\sqrt{1+2x}}\,dx=[\sqrt{1+2x}]_0^1=\sqrt3-1\).

Using the recurrence with \(n=1\):

\(3I_1=\sqrt3-I_0=\sqrt3-(\sqrt3-1)=1\),

so \(I_1=\frac13\).

Using \(n=2\):

\(5I_2=\sqrt3-2I_1=\sqrt3-\frac23\),

hence

\(I_2=\frac{\sqrt3}{5}-\frac{2}{15}\).

Finally, with \(n=3\):

\(7I_3=\sqrt3-3I_2=\sqrt3-3\left(\frac{\sqrt3}{5}-\frac{2}{15}\right)\)

\(=\sqrt3-\frac{3\sqrt3}{5}+\frac{2}{5}=\frac{2\sqrt3+2}{5}\).

So

\(I_3=\frac{2(\sqrt3+1)}{35}\).