I am looking for the antiderivative of $$\frac{\exp(x)-1}{x}$$ I showed that it is equivalent to calculate $$\sum_{n=1}^{\infty}\frac1n \frac{x^n}{n!}$$ but I can't find both of the solutions. If someone could help me I would very appreciate it. Thanks!
2026-03-27 22:52:57.1774651977
Anti-derivative of $\frac{\exp(x)-1}{x}$
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As long as you cannot use the exponential integral function, leave it as you wrote $$\int\frac{e^x-1}{x}\,dx=\sum_{n=1}^{\infty}\frac1n \frac{x^n}{n!}>\sum_{n=1}^{\infty}\frac1{n+1} \frac{x^n}{n!}=\frac 1x\sum_{n=1}^{\infty} \frac{x^{n+1}}{(n+1)!}=\frac{e^x-x-1}x$$