# How do you integrate #int x^2 e^(4-x) dx # using integration by parts?

For integration by parts it's used this scheme:

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To integrate ( \int x^2 e^{4-x} , dx ) using integration by parts:

Let ( u = x^2 ) and ( dv = e^{4-x} , dx ).

Then, ( du = 2x , dx ) and ( v = -e^{4-x} ).

Using the integration by parts formula:

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

Substitute the values:

[ = x^2 (-e^{4-x}) - \int (-e^{4-x}) (2x , dx) ]

[ = -x^2 e^{4-x} + 2 \int x e^{4-x} , dx ]

Now, we perform integration by parts again:

Let ( u = x ) and ( dv = e^{4-x} , dx ).

Then, ( du = dx ) and ( v = -e^{4-x} ).

[ = -x^2 e^{4-x} + 2 \left( -x e^{4-x} - \int -e^{4-x} , dx \right) ]

[ = -x^2 e^{4-x} - 2x e^{4-x} + 2 \int e^{4-x} , dx ]

[ = -x^2 e^{4-x} - 2x e^{4-x} + 2(-e^{4-x}) + C ]

[ = -x^2 e^{4-x} - 2x e^{4-x} - 2e^{4-x} + C ]

Thus, the integration of ( \int x^2 e^{4-x} , dx ) using integration by parts is ( -x^2 e^{4-x} - 2x e^{4-x} - 2e^{4-x} + C ), where ( C ) is the constant of integration.

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