Answer:

Write

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

Use integration by parts with

\(u=(\ln x)^{n-1},\qquad dv=\ln x\,dx.\)

Then

\(du=(n-1)(\ln x)^{n-2}\frac{1}{x}\,dx,\qquad v=x\ln x-x.\)

Therefore

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

The boundary term is zero, since \(x\ln x-x=0\) at \(x=e\), and \((\ln x)^{n-1}=0\) at \(x=1\) for \(n\geq 2\). Hence

\(I_n=-(n-1)\int_1^e\left((\ln x)^{n-1}-(\ln x)^{n-2}\right)dx.\)

So

\(I_n=(n-1)(I_{n-2}-I_{n-1}).\)

Now

\(I_0=e-1,\qquad I_1=\int_1^e\ln x\,dx=1.\)

Using the reduction formula,

\(I_2=I_0-I_1=e-2,\)

and

\(I_3=2(I_1-I_2)=2(1-(e-2))=6-2e.\)

The mean value of \((\ln x)^3\) on \([1,e]\) is

\(\frac{1}{e-1}\int_1^e(\ln x)^3\,dx=\frac{I_3}{e-1}=\frac{6-2e}{e-1}.\)