# How do you find the indefinite integral of #int (-2x^-3+10x^-5) dx#?

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To find the indefinite integral of (\int (-2x^{-3} + 10x^{-5}) , dx), you can use the power rule for integration, which states that (\int x^n , dx = \frac{x^{n+1}}{n+1} + C), where (n \neq -1) and (C) is the constant of integration.

So applying the power rule:

[\int (-2x^{-3} + 10x^{-5}) , dx = -2 \int x^{-3} , dx + 10 \int x^{-5} , dx]

For ( \int x^{-3} , dx), (n = -3), so:

[ \int x^{-3} , dx = \frac{x^{-3+1}}{-3+1} + C = -\frac{1}{2}x^{-2} + C]

For ( \int x^{-5} , dx), (n = -5), so:

[ \int x^{-5} , dx = \frac{x^{-5+1}}{-5+1} + C = -\frac{1}{4}x^{-4} + C]

Therefore, combining the results:

[\int (-2x^{-3} + 10x^{-5}) , dx = -2 \left(-\frac{1}{2}x^{-2}\right) + 10 \left(-\frac{1}{4}x^{-4}\right) + C] [= x^{-2} - \frac{5}{2}x^{-4} + C] [= \frac{1}{x^2} - \frac{5}{2x^4} + C]

So, the indefinite integral of (\int (-2x^{-3} + 10x^{-5}) , dx) is (\frac{1}{x^2} - \frac{5}{2x^4} + 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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