How do you find the integral of #(e^(-1/x)) / x^2#?

1 Answer
Truong-Son N.
Jun 20, 2015

At first, I was thinking why it was written like this. It makes sense... and this is NOT integration by parts. This can be done by u-substitution.

#= int e^"-1/x"/x^2dx = int e^"-1/x"*1/x^2dx#

Let:
#u = e^"-1/x"#
#du = e^"-1/x"*1/x^2dx#

This is rather interesting; there's no need to use #u# here. We have the whole thing as #du#.

#=> int du#

#= u + C#

#= e^"-1/x" + C#

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