How do you integrate #int e^(1/x^2)/x^3dx#?
1 Answer
Jan 7, 2017
Explanation:
#inte^(1//x^2)/x^3dx#
Let
We currently have
#=inte^(1//x^2)(1/x^3dx)=-1/2inte^(1//x^2)(-2/x^3dx)=-1/2inte^udu#
The integral of
#=-1/2e^u=-e^(1//x^2)/2+C#