Question: How can you calculate the integral $$\int\_a^b\frac{\exp(kx)}{x}\,\mathrm dx$$for any $k>0$?

Answer: 
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 $$\int\_{1}^{2}\sum\_{n=-1}^{∞}\frac{2^{n+1}x^{n}}{(1+n)!}dx $$
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
$$\mathrm{Ei}(x)=\int\_{-\infty} ^{x} \dfrac{e^t}{t} \mathrm{d}t$$