Evaluate $\int \frac{1}{\log^2(x)+\log(x)} dx$

Question

I am calculating the following integral:

$\int \frac{1}{\log^2(x)+\log(x)} dx$

Calculating this integral with the substitution method ($\log(x) = t$) I get

$x(\log^2(x)-\log(x)+1)+c$

but Wolfram says it is:

$\operatorname{li}(x) - \dfrac{Ei(\log(x)+1)}{e} + C$.

How would you solve this integral? Is the Wolfram's solution correct? There exists a simpler solution?

Kind regards, Antonio