# How do you integrate #(x^3-4x-10)/(x^2-x-6)# using partial fractions?

The first step is to factor the denominator of the function

since the factors are linear then the coefficients of the partial fractions will be constants , say A and B. Writing the function in terms of it's partial fractions.

multiplying through by (x-3)(x+2)

We now have to find the values of A and B .Note that if x = -2 the term with A will be zero and if x = 3 the term with B will be zero. We can make use of this fact in finding A and B.

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To integrate ((x^3 - 4x - 10)/(x^2 - x - 6)) using partial fractions, follow these steps:

- Factor the denominator (x^2 - x - 6 = (x - 3)(x + 2)).
- Write the given expression as a sum of partial fractions: ((x^3 - 4x - 10)/(x^2 - x - 6) = \frac{A}{x - 3} + \frac{B}{x + 2}).
- Multiply both sides by the denominator ((x^2 - x - 6)) to clear the fractions.
- After multiplying and simplifying, equate the numerators to find the values of (A) and (B).
- Once you find (A) and (B), integrate each term separately.
- The final result will be the sum of the integrals of (\frac{A}{x - 3}) and (\frac{B}{x + 2}).

Let me know if you need further clarification on any step!

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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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