Ex. 2. Resolve \frac{mx+n}{(x-a)(x+b)} into partial fractions.
Assume \frac{mx+n}{(x-a)(x+b)} = {A}{x-a} + {B}{x+b}
mx+n= A(x+b)+B(x-a) (1).
We might now equate coefficients and find the values of A and B, but it is simpler to proceed in the following manner:
Since A and B are independent of x, we may give to x any value we please.
In (1) put x-a=0, or x=a; then A = \frac{ma+n}{a+b};
putting x+b= 0, or x =-b, B = \frac{mb+n}{a+b}
\frac{mx+n}{(x-a)(x+b)} = {ma+n}{x-a} + {mb+n}{x+b}
Ex. 3. Resolve \frac{23x-11x^2}{(2x-1)(9-x^2)} into partial fractions.
Assume \frac{23x-11x^2}{(2x-1)(9-x^2)} = \frac{A}{2x-1} + \frac{B}{3+x} + \frac{C}{3-x}
23x-11x^2 = A(3+x)(3-x) + B(2x-1)(3-x) + C(3+x)(2x-1)
By equating the coefficients of like powers of x, or putting in succession 2x-1=0, 3+x=0, 3-x=0, we find that
A=1, B=4, C=-1. \frac{23x-11x^2}{(2x-1)(9-x^2)} = \frac{1}{2x-1} + \frac{4}{3+x} - \frac{1}{3-x}
Ex. 4. Resolve \frac{3x^2+x-2}{(x-2)^2(1-2x)} into partial fractions.
Assume \frac{3x^2+x-2}{(x-2)^2(1-2x)} = \frac{A}{1-2x} + \frac{B}{x-2} + \frac{C}{(x-2)^2}
3x^2+x-2 = A(x-2)^2 + B(1-2x)(x-2) + C(1-2x)
Let 1-2x =0, then A=-1 3; let x2=0, then C=-4.
To tind B, equate the coefficients of x^2; thus 3 = A-2B; whence B=-5 3.
\frac{3x^2+x-2}{(x-2)^2(1-2x)} = -\frac{1}{3(1-2x)} - \frac{5}{3(x-2)} - \frac{4}{(x-2)^2}
