Answer
The inequality is valid for values less than -1 and values more than 1 (not including them) i.e. $(-\infty,-1)\cap (1,\infty)$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$\dfrac{x+1}{x-1}=0$
$\dfrac{x+1}{x-1}\cdot(x-1)=0(x-1)$
$x+1=0$
$x=-1$
It is important to note that even though we have eliminated the expression $x-1$, the equation has the restriction $x\ne1$; therefore we are going to count it as a critical point.
$x_1=-1$
$x_2=1$
These are the critical points. We are going to take three values: one less than -1; one between -1 and 1; and one more than 1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$\dfrac{-2+1}{-2-1}>0$
$\dfrac{-1}{-3}>0$
$\dfrac{1}{3}>0 \rightarrow \text{ TRUE}$
Second test with a value between -1 and 1:
$\dfrac{0+1}{0-1}>0$
$\dfrac{1}{-1}>0$
$-1>0 \rightarrow \text{ FALSE}$
Third test with a value more than 1:
$\dfrac{2+1}{2-1}>0$
$\dfrac{3}{1}>0$
$3>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x+1}{x-1}>0$ is valid for values less than -1 and values more than 1 (not including them) i.e. $(-\infty,-1)\cap (1,\infty)$
