# How do you integrate #int x / ((x-5)(x+3))dx# using partial fractions?

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To integrate (\frac{x}{(x-5)(x+3)}) using partial fractions, first express the fraction as the sum of two fractions with denominators ((x-5)) and ((x+3)). Then, solve for the unknown constants using algebraic manipulation.

The partial fraction decomposition is:

[\frac{x}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3}]

Multiplying both sides by ((x-5)(x+3)), we get:

[x = A(x+3) + B(x-5)]

Solving for A and B:

[x = A(x+3) + B(x-5)]

[x = Ax + 3A + Bx - 5B]

[x = (A + B)x + (3A - 5B)]

Comparing coefficients:

[1 = A + B]

[0 = 3A - 5B]

Solving the system of equations:

[A + B = 1]

[3A - 5B = 0]

We find (A = \frac{5}{8}) and (B = \frac{3}{8}).

So, the integral becomes:

[\int{\frac{x}{(x-5)(x+3)}}dx = \int{\left(\frac{5}{8}\cdot\frac{1}{x-5} + \frac{3}{8}\cdot\frac{1}{x+3}\right)}dx]

[= \frac{5}{8}\ln|x-5| + \frac{3}{8}\ln|x+3| + C]

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To integrate (\int \frac{x}{(x - 5)(x + 3)} , dx) using partial fractions, you first need to decompose the fraction into partial fractions.

The partial fraction decomposition of (\frac{x}{(x - 5)(x + 3)}) will have the form:

[\frac{x}{(x - 5)(x + 3)} = \frac{A}{x - 5} + \frac{B}{x + 3}]

To find the values of (A) and (B), you can multiply both sides of the equation by the common denominator ((x - 5)(x + 3)) to eliminate the fractions:

[x = A(x + 3) + B(x - 5)]

After expanding and collecting like terms, you'll obtain a linear equation in terms of (x). You can solve this equation to find the values of (A) and (B).

Once you have determined the values of (A) and (B), you can rewrite the original integral as:

[\int \frac{A}{x - 5} + \frac{B}{x + 3} , dx]

This can be integrated term by term, giving you the final result in terms of (A) and (B).

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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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