How do you integrate #(x^2x8)/((x+1)(x^2+5x+6))# using partial fractions?
# int \ (x^2x8)/((x+1)(x^2+5x+6)) \ dx = 3lnx+1 + 2lnx+2 + 2lnx+3 + c #
We seek:
Let's start by breaking down the fraction into its component parts, which will have the following form:
Getting to the answer:
Consequently, we have:
which we can now combine to obtain:
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#= 3ln abs(x+1)+2ln abs(x+2)+2ln abs(x+3) + C#
Take note of this:
So:
So:
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To integrate the function ((x^2  x  8)/((x + 1)(x^2 + 5x + 6))) using partial fractions, follow these steps:

Factor the denominator ((x + 1)(x^2 + 5x + 6)) into linear factors: ((x + 1)(x + 2)(x + 3)).

Write the given fraction as a sum of partial fractions: (\frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}).

Multiply both sides of the equation by the common denominator ((x + 1)(x + 2)(x + 3)) to clear the fractions.

Expand and simplify the equation to solve for (A), (B), and (C).

Once you have found the values of (A), (B), and (C), rewrite the original fraction with the partial fractions.

Integrate each partial fraction separately.

Finally, add the integrals of the partial fractions to find the overall integral of the given function.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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