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How do you evaluate the integral #int e^(1/x)/x^2#?

1 Answer

Jan 28, 2017

#int e^(1/x)/x^2dx = -e^(1/x) + C#

Explanation:

Substitute: #t=1/x#, #dt = -1/x^2dx#

so we have:

#int e^(1/x)/x^2dx = -int e^tdt = -e^t + C #

and undoing the substitution:

#int e^(1/x)/x^2dx = -e^(1/x) + C#

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