How do you use partial fractions to find the integral #int (2x^3-4x^2-15x+5)/(x^2-2x-8)dx#?
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To integrate the given rational function using partial fractions, first factor the denominator. In this case, (x^2 - 2x - 8) factors as ((x - 4)(x + 2)). Then, express the given rational function as the sum of two fractions with undetermined coefficients, one for each factor in the denominator.
The partial fraction decomposition would look like this:
[\frac{2x^3 - 4x^2 - 15x + 5}{x^2 - 2x - 8} = \frac{A}{x - 4} + \frac{B}{x + 2}]
Multiplying both sides by (x^2 - 2x - 8) to clear the denominators, we get:
[2x^3 - 4x^2 - 15x + 5 = A(x + 2) + B(x - 4)]
Next, we can either equate coefficients or choose appropriate values of (x) to solve for (A) and (B). Choosing (x = 4), we eliminate the term containing (B), and choosing (x = -2), we eliminate the term containing (A).
[2(4)^3 - 4(4)^2 - 15(4) + 5 = A(4 + 2) \implies A = \frac{-43}{12}] [2(-2)^3 - 4(-2)^2 - 15(-2) + 5 = B(-2 - 4) \implies B = \frac{13}{12}]
Now, we integrate each term separately:
[\int \frac{-43}{12(x - 4)} + \frac{13}{12(x + 2)} dx]
This integrates to:
[\frac{-43}{12}\ln|x - 4| + \frac{13}{12}\ln|x + 2| + 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.
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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