Integral $\int\frac{1}{x^2\log x}dx.$

Question

I am struggling to evaluate the following integral:

$$ \int\frac{1}{x^2\log x}dx. $$

I have tried to solve it several times, but I wind up with the wrong answer.

Answer 1

Substitute $u = \ln x \implies du=\frac1{x} \, dx$. Note that $x=e^u$.

Now, $$I = \int \frac1{x^2\ln x} \, dx$$ $$=\int \frac{e^{-u}}{u} \, du$$

Note that this is the exponential integral, $-E_1(u)$.