How to find $\int_2^x t/(\log t)^2 \,dt$ $$\int_2^x \frac{t}{(\log t)^2} \,dt,$$
I want to write this integral with $Li(x)$ or $Li_2(x)$. How can i do that?
 A: The integral you wrote has an x in the upper bound of integration, and is the integration variable.  So for now I will give you result of the indefinite integral since it is not clear what you want.$$
\int \frac{x}{\ln^2x} dx= 2Ei(2\ln x)-\frac{x^2}{\ln x}
$$
where Ei is the exponential integral defined by
$$
Ei(x)=-\int_{-x}^\infty \frac{e^{-t}}{t} dt.
$$
A: I'm going to add just a bit more detail.
Let $u = \ln t$.
Then
$$ \int_{2}^{x} \frac{t}{\ln^{2} t} \ dt = \int_{\ln 2}^{\ln x} \frac{e^{u}}{u^{2}} e^{u} \ du = \int_{\ln 2}^{\ln x} \frac{e^{2u}}{u^{2}} \ du$$
Now integrate by parts.
$$ = - \frac{e^{2u}}{u} \Big|^{\ln x}_{\ln 2} + 2 \int_{\ln 2}^{\ln x} \frac{e^{2u}}{u} \ du$$
$$ = -\frac{e^{2 \ln x}}{\ln x} + \frac{e^{2 \ln 2}}{\ln 2} + 2 \int^{\ln x}_{\ln2} \frac{d}{du} \text{Ei} (2u) \ du$$
$$ = - \frac{x^{2}}{\ln x} + \frac{4}{\ln 2} + 2 \Big( \text{Ei}(2 \ln x) - \text{Ei}(2\ln 2) \Big)$$
And assuming $ x >0$,
$$= - \frac{x^{2}}{\ln x} + \frac{4}{\ln 2} + 2 \Big( \text{li}(x^{2}) - \text{li}(4) \Big)$$
where $\text{li}(x)$ is the logarithmic integral.
