Having an integral like $\int_{2}^{10}{\frac{x}{\ln x}}dx$
How does this function turns to an exponential integral of the form:
$ \operatorname{Ei}(x)=-\int_{-x}^{\infty}\frac{e^{-t}}t\,dt.\,$
For instance how did $\int_{2}^{10}{\frac{x}{\ln x}}dx$ became $\text{Ei}\left(2\ln \left(10\right)\right)-\text{Ei}\left(2\ln \left(2\right)\right)$? and how to get a value from $\text{Ei}\left(2\ln \left(10\right)\right)-\text{Ei}\left(2\ln \left(2\right)\right)$?
Is it possible to show in steps how to integrate such function?
For the first part, let $x=e^t$ to make $$\int{\frac{x}{\ln x}}\,dx=\int \frac {e^{2t}}t\,dt=\int \frac {e^{u}}u\,du=\text{Ei}(u)$$
Concerning the evaluation of it, you can use the expansion $$\text{Ei}(u)=\gamma+\log (u) +\sum_{n=1}^\infty \frac {u^n}{n \, n!}$$ and truncate when you consider that you have a sufficient accuracy.
For your case, truncating to $p$ terms, we should have as result $$\left( \begin{array}{cc} p & \text{result} \\ 5 & 22.608730259124323872 \\ 10 & 27.090361747387535417 \\ 15 & 27.158393354747179833 \\ 20 & 27.158556390565134126 \\ 25 & 27.158556489020573262 \\ 30 & 27.158556489040577392 \end{array} \right)$$
For large values of $u$, use the asymptotics $$\text{Ei}(u)\sim\frac{e^u}u \sum_{n=0}^\infty \frac {n!}{u^n}$$ For $u=100$, you should get $$\left( \begin{array}{cc} n & \text{result} \\ 0 & 2.68811714181614\times 10^{41} \\ 1 & 2.71499831323430\times 10^{41} \\ 2 & 2.71553593666266\times 10^{41} \\ 3 & 2.71555206536551\times 10^{41} \\ 4 & 2.71555271051362\times 10^{41} \\ 5 & 2.71555274277103\times 10^{41} \\ 6 & 2.71555274470648\times 10^{41} \\ 7 & 2.71555274484196\times 10^{41} \\ 8 & 2.71555274485279\times 10^{41} \\ 9 & 2.71555274485377\times 10^{41} \\ 10 & 2.71555274485387\times 10^{41} \end{array} \right)$$