Heaviside Method Solution to Problem 11.7-2


© 2001 by Karl Hahn 		KCT logo

This page applies the Heaviside method of partial fractions to find the solution. To see the standard method applied to this same problem, click here.

Even though using the Heaviside method is easier than the standard method, using a calculator is still recommended for this problem.

The problem is to find the indefinite integral 


   


   6x3 + 55x2 + 224x + 1980
                              dx
 (x - 6)(x + 5)(x2 + 4x + 40)

Step 1: Set it up. The integrand is a rational function with a fourth degree polynomial in its denominator. So there will be four unknowns. You can call them, A, B, C, and D, although any four distinct symbols would have been ok. There are no repeated roots in the denominator, but only to of the roots are real (that's because that quadratic can't be factored). So to set it up, just put one of the unknowns over each binomial, and put the other two over the quadratic the way we did in the main text. Then equate the sum of those partial fractions to the integrand. Write that out, and then click here.

You should have 


   6x3 + 55x2 + 224x + 1980
                               =
 (x - 6)(x + 5)(x2 + 4x + 40)


    A       B        Cx + D
        +       +             
  x - 6   x + 5   x2 + 4x + 40

although any ordering of A, B, C, and D over the four binomials listed would be ok.

Step 2: Solve for A by multiplying the above through by  (x - 6)  and evaluating at  x = 6.  Work it through and then click here.

You should have 


   6x3 + 55x2 + 224x + 1980
                            dx  =
    (x + 5)(x2 + 4x + 40)

        B(x - 6)   (Cx + D)(x - 6)
  A  +           +                
          x + 5      x2 + 4x + 40



          1296 + 1980 + 1344 + 1980
       =                             =  6
x = 6              11 × 100

Observe that the B expression and the  Cx + D  expression are both eliminated because you multiplied them by zero. Hence you have established that  A = 6  from this equation alone.

Step 3: Solve for B by multiplying the setup through by  (x + 5)  and evaluating at  x = -5.  Work it through and then click here.

You should have 


   6x3 + 55x2 + 224x + 1980
                            dx  =
    (x - 6)(x2 + 4x + 40)

  A(x + 5)         (Cx + D)(x + 5)
           +  B  +                
    x - 6            x2 + 4x + 40



           -750 + 1375 - 1120 + 1980
        =                             =  -3
x = -5             (-11) × 45

Observe that the A expression and the  Cx + D  expression are both eliminated because you multiplied them by zero. Hence you have established that  B = -3  from this equation alone.

Step 4: Put the two coefficients you have solved for over a common denominator and subtract them out. You have gone as far as you can go with the Heaviside method. This is because you have already solved for the coefficients for all the real roots of the denominator. All that's left are nonreal roots. Subtracting out the known partial fractions will simplify the problem and make finding the remaining coefficients easier.

You know that  A = 6  and  B = -3.  Put those values into the setup equation for A and B. Then put the A and B partial fractions over the common denominator. Then subtract them both from both sides. That will cancel the A and B partial fractions from the right-hand side and will alter the numerator of the left-hand side. Work it all through and then click here.

To get the two partial fractions over a common denominator you have to multiply the A fraction's numerator by  (x + 5)(x2 + 4x + 40) = x3 + 9x2 + 60x + 200.  The B partial fraction's numerator gets multiplied by  (x - 6)(x2 + 4x + 40) = x3 - 2x2 + 16x - 240.  On the left you subtract these products both from its numerator. You know that  A = 6  and  B = -3:  

      6x3  +  55x2  + 224x  +  1980
   -( 6x3  +  54x2  + 360x  +  1200 )      subtracting the A numerator
                                     
                x2  - 136x  +   780
   -(-3x3  +   6x2  -  48x  +   720 )      subtracting the B numerator
                                     
      3x3  -   5x2  -  88x  -    60
The new equation is 


       3x3 - 5x2 - 88x - 60
                                 =
   (x - 6)(x + 5)(x2 + 4x + 40)


      Cx + D
               
   x2 + 4x + 40

Step 5: Divide out both unmatched denominator factors. Both  (x - 6)  and  (x + 5)  appear as factors in the left-hand denominator but do not appear in any denominator on the right. So divide them both out of the left-hand numerator. You will have to do polynomial long division twice, and if you make no mistakes, neither of the divisions can yield any remainder. Do the long divisions, then click here.

You can do the polynomial long divisions in either order. I'm only going to show you one of them here, dividing first by  (x - 6)  and then dividing the result by  (x + 5). 

            3x²  + 13x   - 10    
   x - 6  ) 3x3  -  5x2  - 88x  - 60
            3x3  - 18x2
                       
                   13x2  - 88x
                   13x2  - 78x
                              
                         - 10x  - 60
                         - 10x  - 60
                                    
                                   0
then 
            3x   -  2       
   x + 5  ) 3x2  + 13x  - 10
            3x2  + 15x
                      
                  - 2x  - 10
                  - 2x  - 10
                            
                           0
which means 


       3x3 - 5x2 - 88x - 60
                                 =
   (x - 6)(x + 5)(x2 + 4x + 40)


      3x - 2
                 =
   x2 + 4x + 40


      Cx + D
               
   x2 + 4x + 40

Clearly from the above,  C = 3  and  D = -2. 

Step 6: Put all the solved coefficients for A, B, C, and D into the setup equation, and then integrate. You will have to do both simple substitution and trigonometric substitution to complete this one. See if you can work it through. Then click here.

Here it is 


   


   6x3 + 55x2 + 224x + 1980
                              dx  =
 (x - 6)(x + 5)(x2 + 4x + 40)


  





  6       3
      -       +
x - 6   x + 5


    3x - 2
             
 x2 + 4x + 40






 dx

This one is not as easy to integrate as the first. You have complete the square on the third partial fraction. The substitution for that is 


   v  =  x + 2
and 

v2  =  x2 + 4x + 4
hence 

v2 + 36  =  x2 + 4x + 40

The derived substitutions that you also need for this are  v - 2 = x  and  dv = dx.  


   





  6       3
      -       +
x - 6   x + 5


    3x - 2
             
 x2 + 4x + 40






 dx  =


                    

  6 dx
        -
 x - 6

  

  3 dx
        +
 x + 5

  

 3v - 10
         dx
 v2 + 36

You break the third integral into two integrals. The first of those yields to simple substitution. It's substitution is 



   s  =  v2 + 36
and 

1
  ds  =  v dv
2

The second requires trigonometric substitution. Its substitution is 


              v
   tan(u)  =   
              6
and 


6 (1 + tan2(u)) du  =  dv

With these substitutions you find that 


   





  6       3
      -       +
x - 6   x + 5


    3x - 2
             
 x2 + 4x + 40






 dx  =



   6 ln|x - 6|  -  3 ln|x + 5|  +


  3
   
  2


 ln|x2 + 4x + 40|  -


  4
    arctan
  3



x + 2
     
  6




  +  K




Note that to arrive at this, I've skipped some steps here that we have already covered in previous sections.




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