# How do you integrate #int (2-sqrtx)^5/sqrtx# using substitution?

Substitute:

thus:

and undoing the substitution:

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To integrate (\int (2-\sqrt{x})^5/\sqrt{x}) using substitution, let (u = 2 - \sqrt{x}). Then, (x = (2 - u)^2 = 4 - 4u + u^2), and (du = -\frac{1}{2\sqrt{x}} dx). Substituting these into the integral, we get: (\int (2 - \sqrt{x})^5/\sqrt{x} dx = -2 \int u^5 du). Integrate (u^5) to get (-\frac{1}{6}u^6 + C). Substitute (u = 2 - \sqrt{x}) back in to get the final answer: (-\frac{1}{6}(2 - \sqrt{x})^6 + 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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