PARTIAL FRACTIONS

Partial fractions are of great importance in various mathematical operations and especially in Integral Calculus. The present method of reducing equations to partial fractions is very lengthy and time consuming. By the use of a Vedic sutra you have already come across in division - The Paraavartya Sutra. The following examples will help you to understand how to apply the above sutra to get partial equations in a simple and short manner.

Consider the problem    

x2 + 5x + 3 Eq(1) (x + 2)(x - 7)(x - 4)

This can be written in the form 


                    A     +   B    +    C        Eq(2)
                 (x + 2)   (x - 7)   (x - 4)

where A ,  B , C are constants to be determined.

Now equate each term of the denominator to zero to get three different values of x. Now leaving out the term containing the particular value of x, substitute for x in all the other terms of the given equation to obtain respective constant for the term in the second equation.

i.e.    Equating  (x + 2) to zero we obtain   x + 2 = 0    ==>    x = -2

Now substituting for x in the first equation by (-2) we obtain the constant A.


      Therefore   A  =   (-2)2 + 5(-2) + 3  =  -11
                          (-2 - 7)(-2 - 4)      54

      Similarly   B  =   (7)2 + 5(7) + 3    =   52
                          (7 + 2)(7 - 4)        27

      Similarly   C  =   (4)2 + 5(4) + 3    =   31
                          (4 - 7)(4 + 2)       -18

The above steps can be put in the form of a few simple formulae as follows :


If the equation is of the form :     Lx2 + Mx + N    
                                 (x - a)(x - b)(x - c)
                                 
Then the partial fractions can be written as    
               A    +    B    +    C  
            (x - a)   (x - b)   (x - c)                               
            
 where the constants A, B, C are given as
   A =  La2 + Ma + N     B =  Lb2 + Mb + N      C =  Lc2 + Mc + N
       (a - b)(a - c)        (b - a)(b - c)         (c - a)(c - b)

Example 2 :      Consider


               8x2 + 9x + 11     
          (x - 12)(x + 13)(x + 14)  
          
 This can be written in the form     A     +     B     +     C
                                  (x - 12)    (x + 13)    (x + 14)
                                  
Using the above explained formulas we get
    A = 1271/650   B = -1246/25    C = 1453/26

If one or more of the terms (factors) are in repetition then the problem is solved as follows :

Consider the following example : 


                4x2 + 7x - 3
               (x + 2)2(x - 3)
               
This can be written as    A    +     B    +    C  
                       (x + 2)    (x + 2)2  (x - 3)               
                       
Factorizing and equating the numerators of the two equations, we get

   4x2 + 7x - 3  =  A(x + 2)(x - 3) + B(x - 3) + C(x + 2)
Now substituting the values of x such as to obtain the constants A, B, C i.e. substitute x = 3 and x = -2 such that some terms on the right hand side cancel out to give the required constants. Here the constants are A = 4 , B = 1/5 , C = 54/5.

Another easier method is to equate the coefficients of the x and x2 terms and solve.