A rational expression f(x)/g(x) is called a proper fraction if the degree of f(x) is less than degree of g(x), where g(x) can be factored into linear factors and quadratic factors without real zeros. Now f(x) g(x) can be expressed in simpler terms, namely, as a sum of expressions of the form
How to resolve the given linear fraction as partial fraction ?
First we have to factorize the denominator into prime factors. There are three types in partial fraction.
When the factors of the denominator of the given fraction are all linear factors none of which is repeated. We write the partial fraction as follows.
(x +3)/(x + 1) (x - 2)
here the denominator is in the form linear factors and no factor is repeated. So we can write the partial-fraction as
(x +3) / (x + 1) (x - 2) = [A/(x + 1)] + [B/(x - 2)]
where A and B are constants.
When the factors of the denominator of the given fraction are all linear factors none of which is repeated. We write the partial fraction as follows.
If a linear factor (a x + b) occurs n times as a factor of the denominator of the given fraction, then we can write the partial-fraction as
(x + 3)/(x - 2)³
= [A/(x - 2)] + [B/(x - 2)²] + [C/(x - 2)³]
where A, B and C are constants.
If a quadratic equation a x² + b x + c which is not favorable into linear factors occurs only once as factor of the denominator of the given fraction, then we can write the partial fraction as
Resolve the following rational expressions into partial fractions.
(1) 1/(x^{2}-a^{2}) Solution
(2) (3x + 1)/(x - 2) (x + 1) Solution
(3) x/(x^{2} + 1)(x - 1)(x + 2) Solution
(4) x/(x-1)^{3 }Solution
(5) 1/(x^{4} - 1) Solution
(6) (x - 1)^{2} / (x^{3} + x) Solution
(7) (x^{2} + x + 1)/(x^{2} - 5x + 6) Solution
(8) (x^{3} + 2x + 1)/(x^{2} + 5x + 6) Solution
(9) (x + 12) / (x + 1)^{2} (x - 2) Solution
(12) (7 + x) / (1 + x)(1 + x^{2}) Solution
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