+  < -1

Get 0 on the right by adding 1 to both sides:

 +  + 1 < 0

Th LCD is (x-1)(1+x).  

· + · + 1· < 0

 +  +  < 0

Multiply out the tops but not the bottoms:

 +  +  < 0

 +  +  < 0

Write the sum of the numerators over the LCD:

 < 0

Take away the parentheses on top:

 < 0

 < 0

We find the critical numbers by setting the numerator and
the denominator = 0 and solving:

Numerator = 0

x² + 4x - 1 = 0

That does not factor so we must use the quadratic formula:











Fcator out 2 on the top







 is approximately 0.24 and is approximately -4.24  

That's two of the critical numbers.

Set denominator = 0

(x - 1)(1 + x) = 0

x - 1 = 0;  1 + x = 0
    x = 1;      x = -1

So the four critical numbers are

 which is approximately 0.24 
 which is approximately -4.24
1
-1

We put those in order smallest to largest:

 which is approximately -4.24
-1
 which is approximately 0.24
1

The possible solution intervals are the intervals between and beyond
the critical numbers.  None of the critical numbers are solutions, since
the inequality is < and not < so all the possible intervals are 
open. They are:










We pick a test value in each interval:

, pick test value -5
, pick test value -2
, pick test value 0 
, pick test value .5
 2

We substitute each test value into

 < 0

Substituting test value -5

 < 0
 < 0
 < 0
 < 0
That is false so 
is not part of the solution

Substituting test value -2

 < 0
 < 0
 < 0
 < 0
That is true so  
is part of the solution

Substituting test value 0

 < 0
 < 0
 < 0
1 < 0

That is false so  
is not part of the solution

Substituting test value .5

 < 0
 < 0
 < 0
 < 0
 < 0
That is true so ,  
is part of the solution

Substituting test value 2

 < 0
 < 0
 < 0

That is false so  
is not part of the solution.

So the solution is

 ⋃ 

Edwin