Answer: Show that \(\frac{\mathrm{d}}{\mathrm{d} t}\bigl(t(1+t^{3})^{n}\bigr)=(3n+1)(1+t^{3})^{n}-3n(1+t^{3})^{n-1}\).

For \(I_n=\int_0^1(1+t^3)^n\,\mathrm{d}t\), the reduction formula is \((3n+1)I_n=2^n+3nI_{n-1}\). Hence \(I_3=\frac{319}{140}\).

Differentiate using the product rule:

\(\frac{\mathrm{d}}{\mathrm{d}t}\bigl(t(1+t^3)^n\bigr)=(1+t^3)^n+t\cdot n(1+t^3)^{n-1}\cdot 3t^2\).

So

\(\frac{\mathrm{d}}{\mathrm{d}t}\bigl(t(1+t^3)^n\bigr)=(1+t^3)^n+3nt^3(1+t^3)^{n-1}.\)

Now write \(t^3=(1+t^3)-1\):

\(3nt^3(1+t^3)^{n-1}=3n\bigl((1+t^3)-1\bigr)(1+t^3)^{n-1}\)

\(=3n(1+t^3)^n-3n(1+t^3)^{n-1}.\)

Therefore

\(\frac{\mathrm{d}}{\mathrm{d}t}\bigl(t(1+t^3)^n\bigr)=(3n+1)(1+t^3)^n-3n(1+t^3)^{n-1}.\)

Now integrate from \(0\) to \(1\):

\(\int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}\bigl(t(1+t^3)^n\bigr)\,\mathrm{d}t =\int_0^1\Bigl((3n+1)(1+t^3)^n-3n(1+t^3)^{n-1}\Bigr)\,\mathrm{d}t.\)

The left-hand side is

\(\bigl[t(1+t^3)^n\bigr]_0^1=1\cdot 2^n-0=2^n.\)

The right-hand side is

\((3n+1)I_n-3nI_{n-1}.\)

So

\((3n+1)I_n=2^n+3nI_{n-1}.\)

To find \(I_3\), first calculate \(I_1\):

\(I_1=\int_0^1(1+t^3)\,\mathrm{d}t=\left[t+\frac{t^4}{4}\right]_0^1=1+\frac14=\frac54.\)

Then use the recurrence with \(n=2\):

\(7I_2=2^2+6I_1=4+6\cdot\frac54=4+\frac{30}{4}=\frac{23}{2},\)

hence

\(I_2=\frac{23}{14}.\)

Now use the recurrence with \(n=3\):

\(10I_3=2^3+9I_2=8+9\cdot\frac{23}{14}=8+\frac{207}{14}=\frac{319}{14}.\)

Therefore

\(I_3=\frac{319}{140}.\)