Partial fractions

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Partial fractions

Sometimes it is useful to express a single fraction such as the sum of 2 (or more in other cases) separate fractions.

This is called decomposing a function, f(x), in partial fractions.

Consider this example:

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This equation can be split into the sum of two single fractions.

Therefore

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To find the value of the constants A and B depends on the factors in the denominator. Follow the examples below.

Using the example above:

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Adding the two fractions on the right hand side gives:

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As the denominators are now the same the numerators must match as well. Therefore:

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If we choose to substitute x = -1, this will eliminate the constant B. This gives:

-2 − 1 = A (-3 + 2)
-3 = -A
A = 3

If we choose to substitute x = -2/3 to this will eliminate the constant A. This gives:

-2/3 × 2 − 1 = B (-2/3 + 1)
-4/3 − 1 = B (1/3)
-7/3 = 1/3 B
-7 = B
B = -7

Therefore:

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Lets take the example:

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As this has a quadratic factor in the denominator we need to write the partial equations in the following format:

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Once again we need to add the two fractions on the right hand side. This gives:

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Once again the denominators are now the same, so the numerators must be the same as well. Therefore:

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By substituting x = 0, we eliminate B and C. This gives:

02 + 3 = A (02 + 2)
3 = 2A
A = 3/2

We can substitute two other values for x and solve B and C by simultaneous equations.

Consider this example:

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The denominator contains (x − 2)2. Since, (x − 2)2 = (x − 2 )(x − 2) it is called a repeated factor. When there is a repeated factor in the denominator we need to write the partial equations in the following format:

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Once again we add the fractions on the right hand side together and thus the denominators are the same, so the numerators are the same, giving:

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We again proceed by substitution to obtain the values of the constants.

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