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Mathématiques
postsecondaires





© The scientific sentence. 2010


Mathématiques 5:
Partial fraction decomposition




Calculus II: Integration of Rational Functions
Partial fraction decomposition techniques


1. Definition:

We are interseted in evaluating integrals
of rational functions: R(x) = P(x)/Q(x)

Degree of P(x) > degree of Q(x).

Dividing P(x) by Q(x) gives:

P(x)/ Q(x) = A(x) + B(x)/Q(x)

B(x) is the remainder, and degree of B(x) < degree of Q(x).

Now, we focus on the technique to decompose
P(x)/Q(x) when the degree of P(x) is < degree of Q(x).



2. The technique of partial
fraction decomposition:

2.1. Case1:

If
Q(x) = (a1 x + b1)(a2 x + b2)(a3 x +
b3) ... (an x + bn)

Then:

P(x)/Q(x) = A1/ (a1 x + b ) + A2/ (a2 x +
b2) + A3/ (a3 x + b3) + ... + An/ (an x + bn)

2.2. Case2:

If
Q(x) = (a x + b)n

Then:

P(x)/Q(x) = A1/ (a x + b)1 + A2/ (a x + b)2 +
A3/ (a x + b)3 +... + An/ (a x + b)n

2.3. Case3:

If
Q(x) = (a1 x2 + b1x + c1)(a2 x2 +
b2x + c2)(a3 x2 + b3x + c3) ... (an x2 + bnx +cn)

Then:

P(x)/Q(x) = (A1x + B1)/ (a1 x2 +
b1x + c1) + (A2 x + B2)/ (a2 x2 + b2x + c2) +
(A3x + B3)/ (a3 x2 + b3x +
c3) + ...+ (An x + Bn)/ (an x2 + bnx + cn)

2.4. Case 4:

If
Q(x) = (a x2 + bx + c)n
Then:

P(x)/Q(x) = (A1x + B1)/ (a x2 + bx + c)1 +
(A2 x + B2)/ (a x2 + bx + c)2 +
(A3 x + B3)/ (a x2 + bx + c)3 + ... +
(An x + Bn)/ (a x2 + bx + c)n

3. Examples:

3.1. Example 1:

R(x) = 4x - 3 /(x - 1)(x + 3)2
R(x) = a/(x - 1) + b/(x + 3) + c/(x + 3)2

3.2. Example 2:

R(x) = (4x2 + 1)/(x2 + 2)(x - 1)2(x)3
= (a x + b)/(x2 + 2) + c/(x - 1) + d/(x - 1)2 + e/x + f/x2 + g/x3


4. Example of Finding constants:

R(x) = P(x)/Q(x) = (x - 1 )/(x - 2)(x - 3) =
a/(x - 2) + b/(x - 3).


Equating the numerators of P(x) and the one
of the decomposed fraction, yields a formed equation:

x - 1 = a(x - 3) + b(x - 2)

We have two ways:

1. Sustituting the singular points in the formed equation,

2. Equating the coefficients of each degree of the formed equation.


4.1. Method 1

Singular points: 2 and 3
With 2, we have: 2 - 1 = a(2 - 3) + b(2 - 2) or
a = -1

With 3, we have: 3 - 1 = a(3 - 3) + b(3 - 2) or
b = 2


4.2. Method 2

The formed equation is:
x - 1 = a(x - 3) + b(x - 2)

Developping gives:

x - 1 = ax - 3a - 3a + bx - 2b = (a + b)x -3a - 2b so
1 = a + b, and
-1 = -3a - 2b

Substituting the first in the second yields:

- 1 = - 3(1-b) - 2b = -3 + b, then: b = 2 and a = -1

We write:

(x - 1 )/(x - 2)(x - 3) = -1/(x - 2) + 2/(x - 3)

Therefore, the integral becomes easy to evaluate:

∫ dx (x - 1 )/(x - 2)(x - 3) = -1 ∫ dx /(x - 2) + 2∫ dx /(x - 3) =
- ln(x -2) + 2 ln(x - 3) + constant.


General formula:

If Q(x) = Π(i) Jin(x), (products over i)
then:
P(x)/Q(x) = P(x)/Π Jin(x) = Σ (i,n) Gm(x)/Jin(x)
(sum over i, then sun over n of each i.
n is the degree of each factor Ji(x))

Gm(x) is polynomial of degree equal to degree of Ji(x) - 1

P(x)/Q(x) = P(x)/Π (i) Jin(x)
= Σ (i,n) Gm(x)/Jin(x)
m = i + n


. i goes from 1 to the number of polynomial Ji(x) factors in the denominator,
. n the exponent of Jin(x), goes from 0 to N (N is the given exponent of Jin(x) in the expression of Q(x)),
. m in the subscript of the polynomial numerator in the decomposed fraction. It goes from 1 to i+n.






  


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