How do you integrate #(e^x/x)dx #?

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Answer 1 · 2017-04-12T03:37:39

This is sometimes called the exponential integral:

#inte^x/xdx="Ei"(x)+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+x^2/(2!)+x^3/(3!)+...=sum_(n=0)^oox^n/(n!)#

Then:

#e^x/x=1/x+1+x/(2!)+x^2/(3!)+...=1/x+sum_(n=0)^oox^n/((n+1)!)#

So the antiderivative will be:

#inte^x/xdx=int(1/x+1+x/(2!)+x^2/(3!)+...)dx=ln(absx)+x+x^2/(2*2!)+x^3/(3*3!)+...+C#

#inte^x/xdx=ln(absx)+sum_(n=1)^oox^n/(n*n!)+C#