Question

Evaluate
\[
\int \frac{e^x-1}{x}dx
\]
As an infinite series

Answer 1

Recall that \(\large \displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\). Thus \(\large \displaystyle e^x-1=\sum_{n=1}^{\infty}\frac{x^n}{n!}\) and hence \(\large \displaystyle \frac{e^x-1}{x} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}\),

Therefore, \(\large\displaystyle \int \frac{e^x-1}{x}\,dx = \int\sum_{n=1}^{\infty}\frac{x^{n-1}}{n!}\,dx=\ldots \)

Can you take things from here? :-)