Calculus Question: $\int_2^\infty\frac{\log^3(x-1)}{x^2}dx$ I have just taken calculus quiz but I could not find  $\displaystyle \int_2^\infty\frac{\log^3(x-1)}{x^2}dx$? Any help would be appreciated. Thanks in advance.
EDIT:
Forgot to mention, my tutor gave us hints about this question.


*

*Use Taylor series

*$\displaystyle \zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}$


Those the hints that she gave to us.

By substitutions, the following integrals are equivalent:
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= \int_{1}^{\infty} \, \frac{\log^3(x)}{(1+x)^2} \, dx\\
&= -\int_{0}^{1} \, \frac{\log^3(x)}{(1+x)^2}\, dx \tag 1
\end{align*}
$(1)$ can be written as a sum, consider:
\begin{align*}
  \int_{0}^{1} \, \frac{x^a}{(1+x)^2} dx &= \int_{0}^{1} \, \sum_{n\ge 0} (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} \int_{0}^{1} \, (-1)^n (n+1)\, x^{a+n} \, dx\\
&= \sum_{n\ge 0} (-1)^n \frac{(n+1)}{a+n+1}\tag 2
\end{align*}
Differentiate $(2)$ w.r.t a thrice and set $a=0$, and from $(1)$,
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^3(x-1)}{x^2}\, dx &= 6\, \sum_{n\ge 0}\frac{(-1)^n}{(n+1)^3}\\
&= \frac{9}{2} \zeta{(3)} \approx 5.40925606421817428
\end{align*}
So, a general result looks like:
\begin{align*}
  \int_{2}^{\infty} \, \frac{\log^n(x-1)}{x^2}\, dx &= \left(1-\frac{1}{2^{n-1}}\right)n!\, \zeta{(n)}
\end{align*}
