Calculate the integral $\int \frac{\exp(kx)}{x}dx$ How can you calculate the integral $$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx$$for any $k>0$?
 A: The integral cannot be expressed by elementary functions, instead what one obtain is $$\int_{a}^{b}\frac{e^{kx}}{x}dx=Ei(bk)-Ei(ak)$$ that is, when the integral is defined. So, if I want to calculate the integral $\int_{1}^{2}\frac{e^{2x}}{x}dx$, I use the this result to obtain $$\int_{1}^{2}\frac{e^{2x}}{x}dx=Ei(4)-Ei(2) \approx 14.6766$$ Now, if you would want to approximate the integral then you could try to express the function as a serie. In our case that would be $$\int_{1}^{2}\sum_{n=-1}^{∞}\frac{2^{n+1}x^{n}}{(1+n)!}dx $$
A: If $0$ is in interval $(a,b)$ then integral does not converge. 
In other cases this integral has no elementary derivation. The result in this case is 
$$\mathrm{Ei}(k\cdot b)-\mathrm{Ei}(k\cdot a)$$
where $\mathrm{Ei}(x)$ is function given by
$$\mathrm{Ei}(x)=\int_{-\infty} ^{x} \dfrac{e^t}{t} \mathrm{d}t$$
A: I think you should take a look at Exponential integrals. According to wiki, it is not an elementary function.
For $k>0$ one gets
$\int_a^b\frac{\exp(kx)}{x}\,\mathrm dx = \int_{ka}^{kb}\frac{\exp(x)}{x}\,\mathrm dx = \text{Ei}(kb)-\text{Ei}(ka)$.
