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From Wikipedia

Partial fraction

In algebra, the partial fraction decomposition or partial fraction expansion is a procedure used to reduce the degree of either the numerator or the denominator of a rational function.

In symbols, one can use partial fraction expansion to change a rational function in the form

\frac{f(x)}{g(x)}

where Æ’ and g are polynomials, into a function of the form

\sum_j \frac{f_j(x)}{g_j(x)}

where gj (x) are polynomials that are factors of g(x), and are in general of lower degree. Thus the partial fraction decomposition may be seen as the inverse procedure of the more elementary operation of addition of fractions, that produces a single rational fraction with a numerator and denominator usually of high degree. The full decomposition pushes the reduction as far as it will go: in other words, the factorization of g is used as much as possible. Thus, the outcome of a full partial fraction expansion expresses that function as a sum of fractions, where:

  • the denominator of each term is a power of an irreducible (not factorable) polynomial and
  • the numerator is a polynomial of smaller degree than that irreducible polynomial.To decrease the degree of the numerator directly, the Euclidean algorithm can be used, but in fact if Æ’ already has lower degree than g this isn't helpful.

The main motivation to decompose a rational function into a sum of simpler fractions is that it makes it simpler to perform linear operations on it. Therefore the problem of computing derivatives, antiderivatives, integrals, power series expansions, Fourier series, residues, linear functional transformations, of rational functions can be reduced, via partial fraction decomposition, to making the computation on each single element used in the decomposition. See e.g. partial fractions in integration for an account of the use of the partial fractions in finding antiderivatives. Just which polynomials are irreducible depends on which field of scalars one adopts. Thus if one allows only real numbers, then irreducible polynomials are of degree either 1 or 2. If complex numbers are allowed, only 1st-degree polynomials can be irreducible. If one allows only rational numbers, or a finite field, then some higher-degree polynomials are irreducible.

Basic principles

The basic principles involved are quite simple; it is the algorithmic aspects that require attention in particular cases. On the other hand, the existence of a decomposition of a certain kind is an assumption in practical cases, and the principles should explain which assumptions are justified.

Assume a rational function R(x) = ƒ(x)/g(x) in one indeterminatex has a denominator that factors as

g(x) = P(x)Q(x)

over a fieldK (we can take this to be real numbers, or complex numbers). If P and Q have no common factor, then R may be written as

A/P + B/Q

for some polynomials A(x) and B(x) over K. The existence of such a decomposition is a consequence of the fact that the polynomial ring over K is a principal ideal domain, so that

CP + DQ = 1

for some polynomials C(x) and D(x) (see Bézout's identity).

Using this idea inductively we can write R(x) as a sum with denominators powers of irreducible polynomials. To take this further, if required, write

G(x)/F(x)n

as a sum with denominators powers of F and numerators of degree less than F, plus a possible extra polynomial. This can be done by the Euclidean algorithm, polynomial case. The result is the following theorem:

Therefore when the field K is the complex numbers, we can assume that each pi has degree 1 (by the fundamental theorem of algebra) the numerators will be constant. When K is the real numbers, some of the pi might be quadratic, so in the partial fraction decomposition a quotient of a linear polynomial by a power of a quadratic will occur. This therefore is a case that requires discussion, in the systematic theory of integration (for example in computer algebra).

Procedure

Given two polynomials P(x) and Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n), where the αi are distinct constants and deg&nbsp;P&nbsp;<&nbsp;n, partial fractions are generally obtained by supposing that

\frac{P(x)}{Q(x)} = \frac{c_1}{x-\alpha_1} + \frac{c_2}{x-\alpha_2} + \cdots + \frac{c_n}{x-\alpha_n}

and solving for the ci constants, by substitution, by equating the coefficients of terms involving the powers of x, or otherwise. (This is a variant of the method of undetermined coefficients.)

This approach does not account for several other cases, but can be modified accordingly:

  • If deg&nbsp;P&nbsp;>&nbsp;deg&nbsp;Q, then it is necessary to perform the division
P(x)


From Yahoo Answers

Question:Find the partial fraction decomposition of 3 ____________ 9x^2 + 9x + 2

Answers:3/(9x^2+9x+2) =3/((3x+1)(3x+2)) =A/(3x+1)+B/(3x+2) multiply by original equation by (3x+1)(3x+2); 3=(3x+2)A+(3x+1)B put x= -1/3 3=1*A+0*B A=3 put x= -2/3 3=0*A-B B= -3 therefore, 3/(9x^2+9x+2) = 3/(3x+1)-3/(3x+2)

Question:So far I'm here. A(x+4)^2+B(x)(x+4)+Cx=48 original equation was integral of 48dx/(x(x+4)^2. I got A to be 3 and C to be -12 which I know to be right. I think B equals -3 but I don't know how they got it. Can you guys help with my algebra? How do you solve for B?

Answers:You got down some part of the partial fraction decomposition. Hm... Let me see now... A(x + 8x + 16) + Bx + 4Bx + Cx = 48 Ax + 8Ax + 16A + Bx + 4Bx + Cx = 48 (A + B)x + (8A + 4B + C)x + 16A = 48 A + B = 0 8A + 4B + C = 0 16A = 48 A = 48/16 A = 3 Then, 3 + B = 0 B = -3 8(3) + 4(-3) + C = 0 24 - 12 + C = 0 12 + C = 0 C = -12 Overall, we have 3/x - 3/(x + 4) - 12/(x + 4) . You got that right! Good job! Keep up the good work!

Question:I'm confused and frustrated. :( It just isn't working.

Answers:[x / (x + 1)] dx = since the denominator is a sum of squares it is not factorable, thus partial fraction is not a suitable method in this circumstance; I rather suggest to do as follows: add and subtract 1 at numerator: [(x + 1 - 1)/ (x + 1)] dx = then break it up into: {[(x + 1) / (x + 1)] - [1/ (x + 1)]} dx = [(x + 1) / (x + 1)] dx - [1/ (x + 1)] dx = (x + 1) canceling out, dx - [1/ (x + 1)] dx = the second integral is immediate in that 1/ (x + 1) is the derivative of arctanx: thus dx - [1/ (x + 1)] dx = x - arctanx + c in conclusion, [x / (x + 1)] dx = x - arctanx + c hope it helps Bye!

Question:Partial Fraction Decomposition of a rational fraction (-3x^2+5x-32)/(x-1)(x^2+9)= A/(x-1) + (Bx+C)/(x^2+9) What are A, B, C? Then, evaluate Integral of (-3x^2+5x-32)/(x-1)(x^2+9). Thanks in advance!

Answers:With your first line combine the RHS over a common denominator and compare numerators. -3x^2 + 5x - 32 = A(x^2 + 9) + (Bx + C)(x - 1) This must be an identity true for all x values. x = 1 gives -30 = 10A, A = -3. Then compare coefficients of x^2. -3 = A + B, B = 0 Compare constants. -32 = 9A - C, -32 = -27 - C, C = 5 Putting it all together (-3x^2 + 5x - 32)/(x - 1)(x^2 + 9) = -3/(x - 1) + 5/(x^2 + 9) These are now standard integrals giving -3*ln(x - 1) + 5*arctan(x/3)

From Youtube

Partial Fraction Decomposition :demonstrations.wolfram.com The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries added daily. Many rational functions can be expressed as a sum of simpler fractions. For example, x + 1/(x - 3)(x - 1) can be expressed as the sum A/x - 3 + B/x - 1. To find A and B, Heaviside's method can be used. First, multiply the original fraction by (x - 1), c... Contributed by: Ed Pegg Jr

Partial Fraction Decomposition - Example 4 :Partial Fraction Decomposition - Example 4. Partial Fraction Decomposition - Example 1. In this video, I do a partial fraction decomposition where the denominator factors as a product of LINEAR and QUADRATIC factors.