# How do you integrate #int (2x+1)/((x+8)(x-1)(x-3)) # using partial fractions?

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To integrate the expression (2x+1)/((x+8)(x-1)(x-3)) using partial fractions, we first decompose the rational function into partial fractions.

We start by expressing the rational function as A/(x+8) + B/(x-1) + C/(x-3), where A, B, and C are constants to be determined.

Multiplying both sides by the common denominator (x+8)(x-1)(x-3), we get: 2x + 1 = A(x-1)(x-3) + B(x+8)(x-3) + C(x+8)(x-1).

We can then solve for A, B, and C by equating coefficients of like terms on both sides of the equation.

After finding the values of A, B, and C, we integrate each term separately. The integral of A/(x+8) will be A*ln|x+8|, the integral of B/(x-1) will be B*ln|x-1|, and the integral of C/(x-3) will be C*ln|x-3|.

Finally, we combine the integrals of each term to obtain the overall integral of the original rational function.

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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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