Answer
$\dfrac{7}{(x+1)(x-1)}+\dfrac{8}{(x+1)^{2}}=\dfrac{15x-1}{(x+1)^{2}(x-1)}$
Work Step by Step
$\dfrac{7}{(x+1)(x-1)}+\dfrac{8}{(x+1)^{2}}$
Evaluate the sum of the two rational expressions:
$\dfrac{7}{(x+1)(x-1)}+\dfrac{8}{(x+1)^{2}}=\dfrac{7(x+1)^{2}+8(x+1)(x-1)}{(x+1)^{3}(x-1)}=...$
Take out common factor $x+1$ from the numerator and simplify:
$...=\dfrac{(x+1)[7(x+1)+8(x-1)]}{(x+1)^{3}(x-1)}=\dfrac{7(x+1)+8(x-1)}{(x+1)^{2}(x-1)}=...$
$...=\dfrac{7x+7+8x-8}{(x+1)^{2}(x-1)}=\dfrac{15x-1}{(x+1)^{2}(x-1)}$
