# Solve 1 x 2 </msup> &#x2212;<!-- − -->

Solve $\frac{1}{{x}^{2}-1}⩾\frac{1}{x+1}$
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rideonthebussp
Use the difference of two squares to get
$\frac{1}{\left(x+1\right)\left(x-1\right)}\ge \frac{1}{x+1}$
Case 1: If $x+1>0⇒x>-1$, we have:
$\begin{array}{}\text{(}x\ne -1,1\text{)}& \frac{1}{x-1}\ge 1\end{array}$
The $x\ne -1,1$ comes from the denominator $\left(x+1\right)\left(x-1\right)$, which is undefined when $x=-1,1$.
If $x-1>0⇒x>1$, then multiplying both sides by $x-1$ gives $1\ge x-1⇒x\le 2$. If $x-1<0$, then because $x>-1$, there are no solutions for this case. Hence the intersection of $x>-1,x>1$, and $x\le 2$ imply $1.
Case 2: If $x+1<0⇒x<-1$, we have:
$\begin{array}{}\text{(}x\ne -1,1\text{)}& \frac{1}{x-1}\le 1\end{array}$
$x-1>0$ is not possible in this case because $x<-1$. Thus $x-1<0$, or just $x<-1$, and there are no more conditions.
So the solution to the inequality is $x<-1$, $1 for $x\in \mathbb{R}$.
###### Not exactly what you’re looking for?
Thomas Hubbard
We note on the side that ${x}^{2}\ne 1$ and multiply by ${x}^{2}-1$, so need to distinguish two cases:
$\left\{\begin{array}{l}{x}^{2}-1>0\to 1\ge x-1,\\ {x}^{2}-1<0\to 1\le x-1.\end{array}$
The second case is impossible and we are left with
$x<-1\vee 1
You can also work in systematic way, using a table of sign variations,
$\begin{array}{}& & -1& & 1& & 2\\ x+1& -& 0& +& +& +& +& +\\ x-1& -& -& -& 0& +& +& +\\ 2-x& +& +& +& +& +& 0& -\\ & +& |& -& |& +& 0& -\end{array}$