# How do I find the antiderivative of #x / (x^2 +1)^2#?

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To find the antiderivative of ( \frac{x}{(x^2 + 1)^2} ), you can use a substitution. Let ( u = x^2 + 1 ). Then ( du = 2x , dx ). Rearranging this, ( \frac{du}{2} = x , dx ). Now, rewrite the integral in terms of ( u ):

[ \int \frac{x}{(x^2 + 1)^2} , dx = \int \frac{1}{2} \cdot \frac{1}{u^2} , du ]

This is now a standard integral that can be evaluated as:

[ \int \frac{1}{2u^2} , du = -\frac{1}{2u} + C ]

Substitute back ( u = x^2 + 1 ) to get the final result:

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

where ( C ) is the constant of integration.

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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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