How do you integrate #int e^(1/x^2)/x^3dx#?

1 Answer
Jan 7, 2017

Answer:

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

Explanation:

#inte^(1//x^2)/x^3dx#

Let #u=1/x^2#. Differentiating this, we see that #du=-2/x^3dx#.

We currently have #1/x^3dx# in the integrand, so we need to multiply by #-2# in the integral and #-1/2# to balance this on the exterior of the integral.

#=inte^(1//x^2)(1/x^3dx)=-1/2inte^(1//x^2)(-2/x^3dx)=-1/2inte^udu#

The integral of #e^u# is itself:

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