Question

Resolve $\frac{x^2-x+1}{(x+1)(x-1{)}^2}$ into partial fractions

Answer 1

Solution:

$\frac{x^2-x+1}{(x+1)(x-1{)}^2}$ can be written as $=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(x-1{)}^2}$

$\frac{x^2-x+1}{(x+1)(x-1{)}^2}$ $=\frac{A(x-1{)}^2+B(x^2-1)+C((x+1))}{(x+1)(x-1{)}^2}$

$x^2-x+1=(A+B)x^2+(-2A+C)x+A-B+C$

On comparing coefficients of $x^2,x$ and constant term we get,

$A+B=1$

$-2A+C=-1$

$A-B+C=1$

On solving above relations between $A,B$ and $C$ we get,

$A=\frac{3}{4},B=\frac{1}{4},C=\frac{1}{2}$

$\frac{x^2-x+1}{(x+1)(x-1{)}^2}$ $=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1}{2(x-1{)}^2}$