+26367 There are numbers\( A\) and \(B\) for which
\(\dfrac A{x-1}+\dfrac B{x+1}=\dfrac{x+2}{x^2-1}\)
for every number x\neq\pm1.
Find \(B\).
\(\small{ \begin{array}{|lrcll|} \hline & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{x^2-1} \quad | \quad x^2-1=(x-1)(x+1)\\\\ & \dfrac A{x-1}+\dfrac B{x+1} &=& \dfrac{x+2}{(x-1)(x+1)} \quad | \quad *(x-1)(x+1)\\\\ & \mathbf{ A(x+1) + B(x-1) } &=& \mathbf{ x+2 } \\ \hline 1.~ x=-1:& 0 - 2B &=& -1+2 \\ & -2B &=& 1 \\ & \mathbf{B} &=& \mathbf{ -\dfrac{1}{2} } \\ \hline 2.~ x =1:& 2A+ 0 &=& 1+2 \\ & 2A &=& 3 \\ & \mathbf{A} &=& \mathbf{ \dfrac{3}{2} } \\ \hline \end{array} }\)
