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

Answer 1

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

We are aware of 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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Answer 2

To integrate ( \int x^n e^{x^n} , dx ) using integration by parts, let ( u = x^n ) and ( dv = e^{x^n} , dx ). Then, ( du = n x^{n-1} , dx ) and ( v = \int e^{x^n} , dx ).

This integral for ( v ) doesn't have a simple closed form, but we can represent it using the Exponential Integral function, denoted as ( \text{Ei}(x) ). So, ( v = \text{Ei}(x^n) ).

Now, we apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int x^n e^{x^n} , dx = x^n \text{Ei}(x^n) - \int \text{Ei}(x^n) \cdot n x^{n-1} , dx ]

This integral may not simplify further depending on the context.

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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.

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.

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.

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.

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