# How can you integrate #xe^(x^2) dx# using integration by parts?

Although integrating by parts is not necessary, since it is specified,

The components formula provides us with;

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To integrate (xe^{x^2} , dx) using integration by parts, you can choose (u = x) and (dv = e^{x^2} , dx). Then, differentiate (u) to get (du) and integrate (dv) to get (v). Finally, apply the integration by parts formula: (\int u , dv = uv - \int v , du).

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To integrate ( xe^{x^2} ) using integration by parts, we select ( u = x ) and ( dv = e^{x^2} , dx ). Then, we find ( du ) and ( v ) accordingly.

[ u = x, \quad dv = e^{x^2} , dx ]

[ du = dx, \quad v = \int e^{x^2} , dx ]

The integral ( \int e^{x^2} , dx ) does not have a elementary antiderivative in terms of standard functions, so we cannot integrate it in this form. Therefore, we have to leave ( v ) as ( e^{x^2} ).

Applying the integration by parts formula:

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

[ \int xe^{x^2} , dx = \frac{1}{2}xe^{x^2} - \frac{1}{2} \int e^{x^2} , dx ]

Now, we can rewrite the integral ( \int e^{x^2} , dx ) in terms of ( xe^{x^2} ):

[ \int xe^{x^2} , dx = \frac{1}{2}xe^{x^2} - \frac{1}{2} \int e^{x^2} , dx ]

[ = \frac{1}{2}xe^{x^2} - \frac{1}{2} \cdot \frac{1}{2}e^{x^2} + C ]

[ = \frac{1}{2}xe^{x^2} - \frac{1}{4}e^{x^2} + C ]

Where ( C ) is the constant of integration. Therefore, the integral of ( xe^{x^2} ) using integration by parts is ( \frac{1}{2}xe^{x^2} - \frac{1}{4}e^{x^2} + C ).

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