# How do you integrate #int x arctan x # using integration by parts?

where

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To integrate ( x \arctan(x) ) using integration by parts, you can use the formula:

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

Choose ( u = \arctan(x) ) and ( dv = x , dx ), then differentiate ( u ) and integrate ( dv ) to find ( du ) and ( v ).

[ du = \frac{1}{1 + x^2} , dx ] [ v = \frac{1}{2}x^2 ]

Now apply the formula:

[ \int x \arctan(x) , dx = \frac{1}{2}x^2 \arctan(x) - \int \frac{1}{2}x^2 \frac{1}{1 + x^2} , dx ]

The integral on the right side can be simplified by making a substitution ( u = 1 + x^2 ):

[ \int \frac{1}{2}x^2 \frac{1}{1 + x^2} , dx = \frac{1}{2} \int \frac{1}{u} , du = \frac{1}{2} \ln|u| + C ]

[ = \frac{1}{2} \ln(1 + x^2) + C ]

Therefore, the integral of ( x \arctan(x) ) using integration by parts is:

[ \frac{1}{2}x^2 \arctan(x) - \frac{1}{2} \ln(1 + x^2) + 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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