Solution 2x2−1=2(x−1)(x+1)\frac{2}{x^2-1}=\frac{2}{(x-1)(x+1)}x2−12=(x−1)(x+1)2 Let: 2(x−1)(x+1)=Ax−1+Bx+1\frac{2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}(x−1)(x+1)2=x−1A+x+1B Multiply by (x−1)(x+1)(x-1)(x+1)(x−1)(x+1): 2=A(x+1)+B(x−1)2=A(x+1)+B(x-1)2=A(x+1)+B(x−1) Put x=1x=1x=1: 2=2A⇒A=12=2A\Rightarrow A=12=2A⇒A=1 Put x=−1x=-1x=−1: 2=−2B⇒B=−12=-2B\Rightarrow B=-12=−2B⇒B=−1 Hence: 2x2−1=1x−1−1x+1\boxed{\frac{2}{x^2-1}=\frac{1}{x-1}-\frac{1}{x+1}}x2−12=x−11−x+11