Pl help me to solve the integral...
$\displaystyle \int \frac{1}{x} e^{-2x} dx$
Thanks in advance
To get started on this, we have
$\displaystyle e^{-2x}=1+(-2x)+\frac{(-2x)^2}{2}+\frac{(-2x)^3}{6}+...$
Divide by $\displaystyle x$ to get $\displaystyle \frac{1}{x}+\sum_{n=1}^{\infty}\frac{(-2)^n x^{n-1}}{n!}$
Integrate this term by term (with respect to $\displaystyle x$).