# How do you integrate #((4x^2-1)^2)/x^3dx #?

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To integrate ( \frac{(4x^2 - 1)^2}{x^3} ) with respect to ( x ), we can use polynomial long division to simplify the expression.

First, divide ( 4x^2 - 1 ) by ( x ): [ \frac{4x^2 - 1}{x} = 4x - \frac{1}{x} ]

Now, we have: [ \frac{(4x^2 - 1)^2}{x^3} = \frac{(x(4x - \frac{1}{x}))^2}{x^3} = \frac{x^2(4x - \frac{1}{x})^2}{x^3} = \frac{(4x - \frac{1}{x})^2}{x} ]

Expanding ( (4x - \frac{1}{x})^2 ): [ (4x - \frac{1}{x})^2 = (4x)^2 - 2(4x)(\frac{1}{x}) + (\frac{1}{x})^2 = 16x^2 - \frac{8}{x} + \frac{1}{x^2} ]

So, we have: [ \frac{(4x^2 - 1)^2}{x^3} = \frac{(16x^2 - \frac{8}{x} + \frac{1}{x^2})}{x} = 16x - \frac{8}{x^2} + \frac{1}{x^3} ]

Now, we can integrate each term separately: [ \int 16x , dx = 8x^2 ] [ \int \frac{8}{x^2} , dx = -\frac{8}{x} ] [ \int \frac{1}{x^3} , dx = -\frac{1}{2x^2} ]

Therefore, the integral of ( \frac{(4x^2 - 1)^2}{x^3} ) is: [ \int \frac{(4x^2 - 1)^2}{x^3} , dx = 8x^2 - \frac{8}{x} - \frac{1}{2x^2} + C ] [ = 8x^2 - \frac{8}{x} - \frac{1}{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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