11TH•Mathematics•Partial Fractions•Question 2

Solution

Resolve into partial fractions: $\dfrac{a-b}{(x-a)(x-b)}\,(a\ne b)$

Let:

\frac{a-b}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}\qquad (a\ne b)

Multiply by $(x-a)(x-b)$ :

a-b=A(x-b)+B(x-a)

Put $x=a$ :

a-b=A(a-b)\Rightarrow A=1

Put $x=b$ :

a-b=B(b-a)=-(a-b)\Rightarrow B=-1

Hence:

\boxed{\frac{a-b}{(x-a)(x-b)}=\frac{1}{x-a}-\frac{1}{x-b}}

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