# How do you integrate #(-2x)/((x^2+2x+2)^2) #?

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To integrate ( \frac{-2x}{(x^2 + 2x + 2)^2} ), use the method of substitution. Let ( u = x^2 + 2x + 2 ). Then, ( du = (2x + 2) dx ).

Substitute ( u ) and ( du ) into the integral:

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

This integral can be simplified to:

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

Now, revert back to the original variable ( x ) by substituting ( u = x^2 + 2x + 2 ):

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

So, the integral of ( \frac{-2x}{(x^2 + 2x + 2)^2} ) is ( -\frac{1}{x^2 + 2x + 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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