# How do you find the integral of #xarctanxdx#?

As you can see, there is one element that is simple to integrate and one that is feasible to separate.

This is what I see when I look at it:

which implies Parts Integration.

(One might be tempted to use partial fraction decomposition with the integral here, but there is a simpler method.)

Consequently, the total integral yields:

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To find the integral of (x\arctan(x)), you can use integration by parts. Let (u = \arctan(x)) and (dv = x , dx). Then, differentiate (u) to find (du) and integrate (dv) to find (v). After that, apply the integration by parts formula:

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

Substitute the values of (u), (dv), (du), and (v), and then integrate. You will eventually get the result.

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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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