# How do you use substitution to integrate #(x-2)^5 (x+3)^2 dx#?

(good, now we don't have to expand a 5th order term)

Notice though that Wolfram Alpha would not agree with this answer, which is... odd.

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To integrate the expression ((x-2)^5(x+3)^2) using substitution, you can let (u = x - 2). Then, (x = u + 2), and (dx = du). Substituting these into the expression gives:

[ \int (u)^5(u + 5)^2 du ]

This can be expanded and integrated using standard techniques.

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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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