Solution Let: 4x2−3x+1(x+1)(x−1)2=Ax+1+Bx−1+C(x−1)2\frac{4x^2-3x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}(x+1)(x−1)24x2−3x+1=x+1A+x−1B+(x−1)2C Multiply by (x+1)(x−1)2(x+1)(x-1)^2(x+1)(x−1)2: 4x2−3x+1=A(x−1)2+B(x+1)(x−1)+C(x+1)4x^2-3x+1=A(x-1)^2+B(x+1)(x-1)+C(x+1)4x2−3x+1=A(x−1)2+B(x+1)(x−1)+C(x+1) Solving gives: A=2,B=2,C=1A=2,\quad B=2,\quad C=1A=2,B=2,C=1 Hence: 4x2−3x+1(x+1)(x−1)2=2x+1+2x−1+1(x−1)2\boxed{\frac{4x^2-3x+1}{(x+1)(x-1)^2}=\frac{2}{x+1}+\frac{2}{x-1}+\frac{1}{(x-1)^2}}(x+1)(x−1)24x2−3x+1=x+12+x−12+(x−1)21