# How do you integrate #int (x+4) / (x^2-x-2) dx# using partial fractions?

Write a system of equations:

The partial fraction decomposition is therefore:

Hopefully this helps!

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I got the same answer in a bit of a different approach.

Sometimes if you can do something to simplify the problem, try it.

As an alternative answer, consider the following trick:

so that we get upon cross-multiplying,

thus giving the system of equations

Now we simply get

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User prefers answers without irrelevant information or introduction words.To integrate (\int \frac{x+4}{x^2-x-2} , dx) using partial fractions, we first factor the denominator:

(x^2 - x - 2 = (x - 2)(x + 1)).

Then, we express the integrand as a sum of two fractions:

(\frac{x+4}{x^2-x-2} = \frac{A}{x-2} + \frac{B}{x+1}).

Multiplying both sides by (x^2 - x - 2), we get:

(x + 4 = A(x + 1) + B(x - 2)).

Expanding the right side, we have:

(x + 4 = Ax + A + Bx - 2B).

Matching coefficients, we get the system of equations:

(A + B = 1),

(A - 2B = 4).

Solving this system, we find (A = -3) and (B = 4).

Therefore, the integral becomes:

(\int \frac{x+4}{x^2-x-2} , dx = \int \frac{-3}{x-2} , dx + \int \frac{4}{x+1} , dx).

Integrating both terms separately, we get:

(-3 \ln|x-2| + 4 \ln|x+1| + 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.

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