Ashley Bell

2021-12-13

How do you integrate $\left(\frac{{e}^{x}}{x}\right)dx$ ?

Laura Worden

This is sometimes called the exponential integral:
$\int \frac{{e}^{x}}{x}dx=Ei\left(x\right)+C$
But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for ${e}^{x}$
${e}^{x}=1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\dots =\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{n!}$
Then:
$\frac{{e}^{x}}{x}=\frac{1}{x}+1+\frac{x}{2!}+\frac{{x}^{2}}{3!}+\dots =\frac{1}{x}+\sum _{n=0}^{\mathrm{\infty }}\frac{{x}^{n}}{\left(n+1\right)!}$
So the antiderivative will be:
$\int \frac{{e}^{x}}{x}dx=\int \left(\frac{1}{x}+1+\frac{x}{2!}+\frac{{x}^{2}}{3!}+\dots \right)dx$
$=\mathrm{ln}\left(|x|\right)+x+\frac{{x}^{2}}{2\cdot 2!}+\frac{{x}^{3}}{3\cdot 3!}+\dots +C$
$\int \frac{{e}^{x}}{x}dx=\mathrm{ln}\left(|x|\right)+\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n\cdot n!}+C$

Joseph Lewis

Answer: ${e}^{x}=\frac{{x}^{0}}{0}+\frac{{x}^{1}}{1}+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots$
$=1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots$
$\int \frac{{e}^{x}}{x}dx$
$=\int \frac{1}{x}\left(1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots \right)dx$
$=\int \left(\frac{1}{x}+1+\frac{x}{2}+\frac{{x}^{2}}{3}+\dots \right)dx$
$=\mathrm{log}|x|+x+\frac{{x}^{2}}{{2}^{2}}+\frac{{x}^{3}}{{3}^{3}}$

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