# How do you integrate #xe^(2x)dx#?

I would use Integration by Parts:

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We can use the formula for Integration By Parts (IBP):

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

Then plugging into the IBP formula:

gives us

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To integrate ( xe^{2x} ) with respect to ( x ), we use integration by parts. Let ( u = x ) and ( dv = e^{2x}dx ). Then, ( du = dx ) and ( v = \frac{1}{2}e^{2x} ).

Applying the integration by parts formula: [ \int u dv = uv - \int v du ]

Substituting the values of ( u ), ( dv ), ( du ), and ( v ): [ \int xe^{2x}dx = \frac{1}{2}xe^{2x} - \int \frac{1}{2}e^{2x}dx ]

Now, integrate ( \frac{1}{2}e^{2x} ) with respect to ( x ): [ \int \frac{1}{2}e^{2x}dx = \frac{1}{4}e^{2x} + C ]

Where ( C ) is the constant of integration.

So, putting it all together: [ \int xe^{2x}dx = \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C ]

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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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