How to evaluate this anti-derivative? How to evaluate 
$$ \int \frac{1}{ \ln x} \ \mathrm{d} x, $$
where $\ln x$ denotes the natural logarithm of $x$? 
My effort: 

We note that 
  $$ \int \frac{1}{ \ln x} \ \mathrm{d} x = \int \frac{x}{x \ln x} \ \mathrm{d} x = \int x \frac{ \mathrm{d} }{ \mathrm{d} x } \left( \ln \ln x \right) \ \mathrm{d} x = x \ln \ln x - \int \ln \ln x \ \mathrm{d} x. $$ 

What next? 
Another approach: 

We can also write 
  $$ \int \frac{1}{ \ln x } \ \mathrm{d} x = \frac{x}{\ln x } + \int \frac{1}{ \left( \ln x \right)^2 } \ \mathrm{d} x = \frac{x}{\ln x } + \frac{x}{ \left( \ln x \right)^2 } + \int \frac{ 2 }{ \left( \ln x \right)^3 } \ \mathrm{d} x = \ldots = x \sum_{k=1}^n \left( \ln x \right)^{-k} + n \int \left( \ln x \right)^{-n-1} \ \mathrm{d} x. $$

What next? 
Which one of the above two approaches, if any, is going to lead to a function consisting of finitely many terms comprised of elementary functions, that is, the kinds of solutions that we are used to in calculus courses? 
Or, is there any other way that can lead us to a suitable enough answer? 
 A: I = $\large\int\frac{1}{ln(x)}dx$
let ln(x) =  u
$\,e^u = x$
$\,dx = e^udu$
I = $\,\int \frac{e^u}{u}du$
expanding e$^u$,
I=$\,\int\frac{1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\frac{u^5}{5!}..............}{u}du$
I = $\,\int\frac{1}{u}+1+\frac{u}{2!}+\frac{u^2}{3!}+\frac{u^3}{4!}+\frac{u^4}{5!}...........du$
I = $\,ln(u)+ u + \frac{u^2}{2.2!}+\frac{u^3}{3.3!}+\frac{u^4}{4.4!}+\frac{u^5}{5.5!}......+ C$
I = $\,ln(ln(x)) + ln(x) + \frac{(ln(x))^2}{2.2!} + \frac{(ln(x))^3}{3.3!} + \frac{(ln(x))^4}{4.4!} + \frac{(ln(x))^5}{5.5!} + ....... + C $
I = $\,ln(ln(x)) + ln(x) + \sum_{k=2}^\infty \frac{(ln(x))^k}{k.k!} + C $
A: It cannot be expressed with the elementary functions. Actually, it is a special function, the integral logarithm:
$$\operatorname{li}(x)=\int_0^x\frac{\mathrm dt}{\ln t}\biggl(=\lim_{\varepsilon\to 0}\int_0^{1-\varepsilon}\frac{\mathrm dt}{\ln t}+\lim_{\varepsilon\to 0}\int_{1+\varepsilon}^\infty\frac{\mathrm dt}{\ln t}\quad\text{for }x>1\biggr)$$
and  it is asymptotically equivalent to $\;\dfrac x{\ln x}$.
Closely linked is the function $$\operatorname{Li}(x)=\int_2^x\frac{\mathrm dt}{\ln t}=\operatorname{li}(x)-\operatorname{li}(2), $$
 used in number theory: it is a asymptotic equivalent of $\pi(x)$, the number of primes $\le x$.
