why is the integral of $e^{-x}/x = (x-1)e^{x}$?

Question

Why is $\int \frac{e^{-x}}{x}\; dx= (x-1)e^x$?

This is the answer in my calculus textbook, I thought that this was the form for an Exponential integral (which I know little about).

Therefore shouldn't the answer be something like:

$-E_1(x) + C$ ?

Answer 1

You are correct. You can easily check on your own that the provided answer is incorrect by noticing that $$\frac{d}{dx}(x-1)e^x=xe^x$$ which you can check by e.g. using the product rule, and this is obviously not the same as $e^{-x}/x$. The correct answer should be $$\int \frac{e^{-x}}{x}dx=\operatorname{Ei}(-x)+C$$ for constant $C$, where $\operatorname{Ei}$ is the exponential integral.