Answer
$\dfrac{4}{5(x-1)}, x\ne-1, 0, 1$
Work Step by Step
The initial restrictions are $x\ne\pm1$.
Use the rule $\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \cdot \frac{d}{c}$ to obtain:
$=\dfrac{8x}{x^2-1} \cdot \dfrac{x+1}{10x}$
This yields an additional restriction $x\ne0$.
Factor each polynomial completely.
Use special formula $a^2+b^2=(a+b)(a-b)$ to obtain:
$=\dfrac{4\cdot 2x}{(x+1)(x-1)} \cdot \dfrac{x+1}{5\cdot 2x}, x\ne0, \pm1$
Cancel the common factor $x+1$:
$=\dfrac{4}{5(x-1)}, x\ne0, \pm1$
Hence, the lowest term is:
$\dfrac{4}{5(x-1)}, x\ne-1, 0, 1$
