How do you integrate #inte^(1/x)/x^2# using substitution?

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Answer 1 · 2017-02-16T08:24:36

#-e^(1/x)+C#

#I = int e^(1/x)/x^2 dx#

Let #u=1/x -> (du)/dx=-1/x^2#

#:. dx=-x^2du#

#I = -int e^u/x^2 * x^2 du#

# = -int e^u/ cancel (x^2) * cancel(x^2) du#

#=-e^u + C = -e^(1/x)+C#

Answer 2 · 2017-02-16T08:24:52

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

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

#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#