Home > Calculus > Integration by Substitution

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

Answer 1

Paisley Johnson

#-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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

William Abdalla

#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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Cameron Alexander

To integrate ( \frac{e^{1/x}}{x^2} ) using substitution, let ( u = \frac{1}{x} ). Then ( du = -\frac{1}{x^2}dx ).

Substitute ( u ) and ( du ) into the integral:

[ \int \frac{e^{1/x}}{x^2} dx = -\int e^u du ]

Now integrate ( -e^u ):

[ -\int e^u du = -e^u + C ]

Finally, substitute back ( u = \frac{1}{x} ) to get the final result:

[ \int \frac{e^{1/x}}{x^2} dx = -e^{1/x} + C ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.