Calculating the integral of $\log e$ with base of $x$ Is it possible to calculate the integral of $\log e$ with base of $x$?
 A: $$
\int\log_x(e)\,\mathrm{d}x=\int\frac1{\log(x)}\,\mathrm{d}x
$$
This is known as the Log-Integral.
$$
\begin{align}
\mathrm{li}(x)
&=\int_0^x\frac1{\log(t)}\,\mathrm{d}t\\[6pt]
&=\lim_{a\to0^+}\int_0^{1-a}\frac1{\log(t)}\,\mathrm{d}t
+\int_{1+a}^x\frac1{\log(t)}\,\mathrm{d}t\\[6pt]
&=\lim_{a\to0^+}\int_{-\infty}^{\log(1-a)}e^s\frac{\mathrm{d}s}{s}
+\int_{\log(1+a)}^{\log(x)}e^s\frac{\mathrm{d}s}{s}\\[6pt]
&=\lim_{a\to0^+}\log|\log(1-a)|\,e^{\log(1-a)}-\int_{-\infty}^{\log(1-a)}\log|s|\,e^s\,\mathrm{d}s\\
&\hphantom{\lim_{a\to0^+}}+\int_{\log(1+a)}^{\log(x)}\frac{\mathrm{d}s}{s}
+\int_{\log(1+a)}^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt]
&=\lim_{a\to0^+}\log|\log(1-a)|\,(1-a)-(-\gamma)+\log|\log(x)|-\log|\log(1+a)|\\
&\hphantom{\lim_{a\to0^+}}+\int_0^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt]
&=\gamma+\log|\log(x)|+\int_0^{\log(x)}(e^s-1)\frac{\mathrm{d}s}{s}\\[6pt]
&=\gamma+\log|\log(x)|+\sum_{k=1}^\infty\frac{\log(x)^k}{k\,k!}
\end{align}
$$
where $\gamma$ is the Euler-Mascheroni Constant.
A: A guess: Since this is reported in comments below the question to have been in a "question sheet" in a course, is it possible that something like the following happened?
The question sheet says "Find the derivative $f'(w)$ if $f(w)=$
#1 etc. etc. etc.
#2 etc. etc. etc.
#3 etc. etc. etc.
#4 ${}\qquad\displaystyle \int_1^w (\log_x e)\,dx$
#5 etc. etc. etc."
Often students lose sight of the words at the beginning and mistakenly think they're being asked to find the integral.
postscript: ($f'(w)$ would of course be $\log_w e$.)
