Lecture in Integration by Partial Fractions

(Last Updated On: December 21, 2017)
Review in Integration by Partial Fractions

In this topic you are going learn how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that is simplified and easy for us to integrate. The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition. It is important algebraic technique because many integrals involving rational expressions can be done if we first do partial fractions on the integrand.

This method is based on the simple concept of adding fractions by getting a common denominator.

The Partial Fractions decomposition for ¾ is

 

This concept can also be used with functions of $ x $ . For example,

$ displaystyle{ {2 over x+1} - {1 over x} } = displaystyle{ {2 over x+1}Big({x over x}Big) - {1 over x}Big({x+1 over x+1}Big) } $
$ = displaystyle{ {2x over x(x+1) } - {x+1 over x(x+1) } } $
$ = displaystyle{ 2x-x-1 over x^2+x } $
$ = displaystyle{ x-1 over x^2+x } $

so that we can now say that a partial fractions decomposition for $  displaystyle{ x-1 over x^2+x }  $ is

$ displaystyle{ x-1 over x^2+x } = displaystyle{ {2 over x+1} - {1 over x} } $

Integrating the function:

= –   = 2ln |x+1| – lnx + C

METHOD: In general Consider a rational function:

where P and Q are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q: Such a rational function is called proper. If f is improper, then we must take the preliminary step of dividing Q into P (by long division) until a remainder R(x) is obtained such that deg(R) < deg(Q): The division statement is

where S and R are also polynomials. Sometimes this preliminary step is all that is required to get the integral.
Example 1. Evaluate
SOLUTION: Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write

= =

The next step is to factor the denominator Q(x) as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of the form ax2 + bx + c; where b2 – 4ac < 0). For instance,

  • x2 – 1 = (x – 1) (x + 1)
  • 10×2 – 11x – 6 = (2x – 3) (5x + 2)
  • x3 + 1 = (x + 1) (x2 – x + 1) 
  • x3 + 5 = (x + 5⅓) (x2 – 5⅓x + 25⅓)
  • x3 + x2 + x + 1 = x3 (x + 1) + x + 1 = (x + 1) (x3 + 1) 
  • 3×3 + 14×2 + 7x – 4 = 3×2 (x + 1) + 11x(x + 1) – 4 (x + 1) = (x + 1) (3x – 1) (x + 4) 
  • x4 – 16 = (x2)2 – 42 = (x2 – 4) (x2 + 4) = (x – 2) (x + 2) (x2 + 4) 
  • x4 + 16 = x4 + 8×2 + 16 – 8×2 = (x2 + 4)2 – 8×2 = (x2 + 2√2x + 4) (x2 – 2 √2x + 4) 
  • x5 + 1 = (x + 1)(x4 – x3 + x2 – x + 1) = ¼ (x + 1) (2×2 – x + √5x + 2) (2×2 – x – √5x + 2) 
  • x6 – 3×5 + 10×3 – 15×2 + 9x – 2 = (x + 2) (x – 1)5
  • x9 + 6×8 + 21×7 + 51×6 + 81×5 + 87×4 + 32×3 – 63×2 – 108x – 108 = (x – 1) (x + 2)2 (x2 + x + 3)3
  • x2 + 1; x2 + 2x + 5; etc. .. are irreducible quadratic factors

The third step is to express the proper rational function R(x)/Q(x) as a sum of partial fractions of the form

or

A theorem in algebra guarantees that it is always possible to do this. There are four possible cases:

Case I: The denominator Q(x) is a product of distinct linear factors.

This means that we can write

Q(x) = (a1x + b1)(a2x + b2)  .  .  .  (akx + bk)

where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1;A2; .  .  . ;Ak such that

Case I: The denominator Q(x) is a product of distinct linear factors.

Example 2: Consider
SOLUTION: Since the degree of the numerator is less than the degree of the denominator,
we don’t need to divide. We factor the denominator as

2×3 + 3×2 – 2x = x( 2×3 + 3x – 2) = x ( 2x – 1 ) (x + 2 )

As you can see, the denominator has three distinct linear factors (with yellow background), therefore, the partial fraction decomposition of the integrand has the form

To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators,x ( 2x – 1 ) (x + 2 ) , obtaining

x2 + 2x – 1 = A (2x – 1) ( x + 2) + Bx ( x + 2) + Cx ( 2x – 1)

Expanding the right side of Equation  and writing it in the standard form for polynomials, we get

x2 + 2x – 1 = (2A + B + 2C) x2 + (3A + 2B) – C) x – 2A

The polynomials in Equation are identical, so their coefficients must be equal. The coefficient of x2 on the right side, 2A + B + 2C, must be equal to the coefficient of x2 on the left side—which is, 1 . Likewise, the coefficients of x are equal and the constant terms are equal. This gives the following system of equations for A,  B, and C:

2A + B + 2C = 1

3A + 2B – C = 2

-2A = -1

Solving, we get A = ½ , B = ⅕ , C = -1/10, integrate the partial fractions

= =

At this point, it is a must that you can already do simple integration in mind. In integrating the middle term you can made the mental substitution u = 2x – 1, which gives du = 2 dx and dx = ½ du

Note: Another way of finding the coefficients of A, B and C is elimination or simplifying the equation

x2 + 2x – 1 = A (2x – 1) ( x + 2) + Bx ( x + 2) + Cx ( 2x – 1)

Let x = 0, the second and third term eliminate, the equation becomes -1 = -2A, therefore A = 1/2
Let x = 1/2, the first and third terms eliminate, 1/4 = 5/4B, then B = 1/5
Let x = -2, the first and second terms eliminate, -1 = 10C, we get C = -1/10

Important: After getting the coefficients of A, B, and C, substitute then  proceed integrating the partial fractions. As you can see this method of getting the coefficients is much simpler and easy compare to the previous one but it is up to you what do you prefer.

Case II: The denominator Q(x) is a product of linear factors, some of which are repeated.

This means that we can write

Case II: The denominator Q(x) is a product of linear factors, some of which are repeated.

Instead of the single term A1/(a1x + b1x) in (Case I), we would use

The form of partial fractions in Case II

Example 3: Find

Solution: The Partial Fraction decomposition is

=

Putting the right side over the common denominator (x – 1)3, you have

=

Equating numerators, (I’m going to use the second method then the first method.)

3×2 – x + 4 = A(x – 1)2 + B(x – 1) + C

Let x = 1, first and second terms eliminate, equation equal to, 3 -1 + 4 = C, thus, C = 6
Equating coefficients of x2 : 3 = A
Equating coefficients of x:   -1 = -2A + B ; B =  -1 + 2(3) ; thus B = 5

Evaluate:

= =

Case III: The denominator Q(x) contains irreducible quadratic factors, none of which is repeated.

For example: x2 + 1;  x(2x – 3) (x2 + x + 1);  x2(x2 + 4) (x2 + 3x + 8); etc.

In this case if Q(x) has the factor ax2 +bx + c; where b2 – 4ac < 0; then, in addition to the partial fractions in (Case I) and (Case II), the expression for R(x)/Q(x) will have a term of the form

, where A and B are constants to be determined.

Example 4: Evaluate

SOLUTION: Factored the denominator, x3 + 4x = x (x2 + 4 ) thus, the partial fraction decomposition is

Multiplying by x (x2 + 4), we have

2×2 – x + 4 = A (x2 + 4) + (Bx + C) x
                  = (A + B)x2 + Cx + 4A

Equating these coefficients
Coefficients of x2 : 2 = (A + B)
Coefficients of x : -1 = C
Coefficients of x0 : 4 = 4A
Then, from 3rd equation A = 1 , from 1st equation using A = 1 we got B = 1 ,  and from the second equation we have C = -1

Hence,

  = = { + – }
=

Important: To integrate the second term I split it into two parts. Make the u-substitution for the second integral so that du = 2xdx. Evaluate the third integral using the Inverse Trigonometric forms of tangent. Trigonometric Table/Formulas.

Case IV: The denominator Q(x) contains a repeated irreducible quadratic factor.

For example: (x2 + x + 2)2; (2x + 7)(x2 + 4)3; x(x – 2)(x + 1)2(2×2 + 3)(x2 + 1)3(x2 + 4x + 5)4; etc.

In this case if Q(x) has the factor (ax2 + bx + c)r; where b2 – 4ac < 0; then, instead of a single partial fraction (In Case III), the sum

Case IV: The denominator Q(x) contains a repeated irreducible quadratic factor.

occurs in the partial fraction decomposition of R(x)/Q(x). Each of the terms in the decomposition can be integrated by using a substitution or by first completing the square if necessary.

Example 5:

SOLUTION: The form of the partial fraction decomposition is

Then,

-x3 + 2×2 – x + 1 = A (x2 + 1)2 + ( Bx + C ) x ( x2 + 1) + ( Dx + E ) x
                                            = A( x4 + 2×2 + 1) + (B x4 + x2) + C( x3 + x ) + Dx2 + Ex
                                      = ( A + B) x4 + Cx3 + ( 2A + B + D )x2 + ( C + E )x + A

Equating the coefficients,
Coefficients of x4: 0 = A + B
Coefficients of x3: C = -1
Coefficients of x2: 2 = 2A + B + D
Coefficients of x1: -1 = C + E
Coefficients of x0: 1 = A
Then solving we get, A = 1 , B = -1 , C = -1 , D = 1 , and E = 0

Evaluate:

=
   =

Important:Again, you will notice that I split the second integral. Try some practice exercises to familiarize yourself to evaluate integral using this method. Integration by Partial Fractions – Set 1 Problems

credit: D. A. Kouba (UC Davis), Kiryl Tsishchanka, James Stewart©2013 www.PinoyBIX.com

Lecture in Integration by Partial Fractions
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