How do you integrate #int x^n*e^(x^n)dx# using integration by parts?

1 Answer
Cesareo R.
Nov 19, 2016

#int x^n e^(x^n)dx= 1/nx e^(x^n)+(x Gamma(1/n, -x^n))/(n (-x^n)^(1/n) )+C#

Explanation:

We know that

#d/dx(xe^(x^n))=nx^n e^(x^n)+e^(x^n)# so

#int x^n e^(x^n)dx = 1/nx e^(x^n)-int e^(x^n)dx#

The integral #int e^(x^n)dx# is a manual integral and is equal to

#int e^(x^n)dx=-(x Gamma(1/n, -x^n))/(n (-x^n)^(1/n) )+C# so

#int x^n e^(x^n)dx= 1/nx e^(x^n)+(x Gamma(1/n, -x^n))/(n (-x^n)^(1/n) )+C#

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