Answer: We have \(I_n=\int_1^e x(\ln x)^n\,dx\) for \(n\ge 1\). Then

\(I_{n+1}=\int_1^e x(\ln x)^{n+1}\,dx\).

Integrate by parts with \(u=(\ln x)^{n+1}\) and \(dv=x\,dx\). Then \(du=(n+1)(\ln x)^n\frac{1}{x}\,dx\) and \(v=\frac{x^2}{2}\).

So

\(I_{n+1}=\left[\frac{x^2}{2}(\ln x)^{n+1}\right]_1^e-\int_1^e \frac{x^2}{2}(n+1)(\ln x)^n\frac{1}{x}\,dx\)

\(=\left[\frac{x^2}{2}(\ln x)^{n+1}\right]_1^e-\frac{n+1}{2}\int_1^e x(\ln x)^n\,dx\).

At \(x=e\), \(\ln e=1\), so the boundary term is \(\frac{e^2}{2}\). At \(x=1\), \(\ln 1=0\), so the boundary term is \(0\). Hence

\(I_{n+1}=\frac{1}{2}e^2-\frac{n+1}{2}I_n\).

Now define the induction statement \(H_n\): \(I_n=A_ne^2+B_n\), where \(A_n\) and \(B_n\) are rational.

First, for \(n=1\),

\(I_1=\int_1^e x\ln x\,dx\).

Using the recurrence with \(n=1\) in the form \(I_2=\frac{1}{2}e^2-I_1\) is not needed; instead evaluate directly by parts:

\(I_1=\left[\frac{x^2}{2}\ln x\right]_1^e-\int_1^e \frac{x^2}{2}\cdot \frac{1}{x}\,dx=\frac{e^2}{2}-\frac{1}{2}\int_1^e x\,dx\)

\(=\frac{e^2}{2}-\frac{1}{2}\left[\frac{x^2}{2}\right]_1^e=\frac{e^2}{2}-\frac{1}{2}\cdot\frac{e^2-1}{2}=\frac{e^2}{4}+\frac14\).

So \(H_1\) is true with \(A_1=\frac14\) and \(B_1=\frac14\).

Assume \(H_k\) is true for some \(k\ge 1\), so \(I_k=A_ke^2+B_k\) with \(A_k,B_k\) rational. Then

\(I_{k+1}=\frac{1}{2}e^2-\frac{k+1}{2}I_k=\frac{1}{2}e^2-\frac{k+1}{2}(A_ke^2+B_k)\)

\(=\left(\frac12-\frac{k+1}{2}A_k\right)e^2-\frac{k+1}{2}B_k\).

Therefore \(I_{k+1}=A_{k+1}e^2+B_{k+1}\), where \(A_{k+1}=\frac12-\frac{k+1}{2}A_k\) and \(B_{k+1}=-\frac{k+1}{2}B_k\). Since rational numbers are closed under these operations, \(A_{k+1}\) and \(B_{k+1}\) are rational.

By induction, for every positive integer \(n\), \(I_n\) is of the form \(A_ne^2+B_n\) with \(A_n\) and \(B_n\) rational.

First derive the recurrence. For \(n\ge 1\),

\(I_{n+1}=\int_1^e x(\ln x)^{n+1}\,dx\).

Take \(u=(\ln x)^{n+1}\) and \(dv=x\,dx\). Then \(du=(n+1)(\ln x)^n\frac{1}{x}\,dx\) and \(v=\frac{x^2}{2}\). Integration by parts gives

\(I_{n+1}=\left[\frac{x^2}{2}(\ln x)^{n+1}\right]_1^e-\int_1^e \frac{x^2}{2}(n+1)(\ln x)^n\frac{1}{x}\,dx\)

\(=\left[\frac{x^2}{2}(\ln x)^{n+1}\right]_1^e-\frac{n+1}{2}\int_1^e x(\ln x)^n\,dx\).

Now \(\ln e=1\) and \(\ln 1=0\), so

\(\left[\frac{x^2}{2}(\ln x)^{n+1}\right]_1^e=\frac{e^2}{2}\).

Hence

\(I_{n+1}=\frac{e^2}{2}-\frac{n+1}{2}I_n\).

Now prove by induction that \(I_n\) has the form \(A_ne^2+B_n\) with \(A_n,B_n\in\mathbb{Q}\).

Let \(H_n\) be the statement \(I_n=A_ne^2+B_n\), where \(A_n\) and \(B_n\) are rational.

Base case: for \(n=1\),

\(I_1=\int_1^e x\ln x\,dx\).

Integrating by parts again with \(u=\ln x\), \(dv=x\,dx\), we get

\(I_1=\left[\frac{x^2}{2}\ln x\right]_1^e-\int_1^e \frac{x^2}{2}\cdot\frac1x\,dx\)

\(=\frac{e^2}{2}-\frac12\int_1^e x\,dx=\frac{e^2}{2}-\frac12\left[\frac{x^2}{2}\right]_1^e\)

\(=\frac{e^2}{2}-\frac{e^2-1}{4}=\frac14e^2+\frac14\).

So \(H_1\) is true with \(A_1=\frac14\) and \(B_1=\frac14\).

Inductive step: assume \(H_k\) is true for some \(k\ge 1\), so \(I_k=A_ke^2+B_k\) with \(A_k,B_k\in\mathbb{Q}\). Then

\(I_{k+1}=\frac{e^2}{2}-\frac{k+1}{2}I_k\)

\(=\frac{e^2}{2}-\frac{k+1}{2}(A_ke^2+B_k)\)

\(=\left(\frac12-\frac{k+1}{2}A_k\right)e^2-\frac{k+1}{2}B_k\).

Thus \(I_{k+1}=A_{k+1}e^2+B_{k+1}\), where

\(A_{k+1}=\frac12-\frac{k+1}{2}A_k\), \(B_{k+1}=-\frac{k+1}{2}B_k\).

Since \(A_k\) and \(B_k\) are rational, it follows that \(A_{k+1}\) and \(B_{k+1}\) are rational.

Therefore, by induction, for every positive integer \(n\), \(I_n\) is of the form \(A_ne^2+B_n\), with \(A_n\) and \(B_n\) rational.