- #1

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It would be best if I show you the examples in the book :

[tex] \frac{3x^2 +12x +11} {(x+1)(x+2)(x+3)} [/tex]

To express this fraction in the form

[tex] \frac{A} {x+1} + \frac{B} {x+2} + \frac{C} {x+3} [/tex]

We have to find the values of A , B and C by solving three simultaneous equations ( by the usual method )

The method proposed in the book :

equate the denominator of [tex]A[/tex] to zero and thus get the 'paravartya'

Then substitute this value is the original equation but without the factor which is A's denominator. The result is the value of A.

so for A, the 'paravartya' is -1 ( from x+1 = 0 )

substituting this in

[tex] \frac {3x^2 +12x +11} {(x+2)(x+3)} [/tex]

we get 1, so A = 1

similarly for B, the 'paravartya' is -2 (from x+2 =0)

substituting this in

[tex] \frac{3x^2 +12x +11} {(x+1)(x+3)} [/tex]

we get 1, so B = 1

For C, the paravartya is -3 ( from x+3 =0)

substituting this in

[tex] \frac{3x^2 +12x +11} { (x+1)(x+2) }[/tex]

we get 1, so C = 1

So, the partial fraction is

[tex] \frac{1} {x+1} + \frac{1} {x+2} + \frac{1} {x+3} [/tex]

This makes integration so much easier!