# How do you find the indefinite integral of #int (3x^-2-4x^-3)dx#?

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To find the indefinite integral of ( \int (3x^{-2} - 4x^{-3}) , dx ), you can use the power rule for integration. The power rule states that for ( \int x^n , dx ), where ( n \neq -1 ), the result is ( \frac{x^{n+1}}{n+1} + C ), where ( C ) is the constant of integration.

Applying the power rule to each term:

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

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

Combining both results, the indefinite integral is: [ \int (3x^{-2} - 4x^{-3}) , dx = -3x^{-1} + 4x^{-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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