Maths Question:
$\text{Resolve into partial fraction}\frac{1}{{{x}^{2}}-1}$
Maths Solution:
$\begin{align} & \frac{1}{{{x}^{2}}-1}=\frac{1}{(x-1)(x+1)} \\ & \frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1} \\ & 1=A(x-1)+B(x+1) \\ & \text{Set }x=-1 \\ & 1=-2A \\ & A=-\frac{1}{2} \\ & \text{Set }x=1 \\ & 1=2B \\ & B=\frac{1}{2} \\ & \frac{1}{(x-1)(x+1)}=\frac{-\tfrac{1}{2}}{x-1}+\frac{\tfrac{1}{2}}{x+1} \\ & \frac{1}{(x-1)(x+1)}=\frac{1}{2}\left( \frac{-1}{x-1}+\frac{1}{x+1} \right) \\\end{align}$
University mathstopic: