To solve (x^2-1)/(x^2-4) > -1

Solution:

(x^2-1)/(x^2-4) +1 > 1

{x^2-1+(x^2-4)}/(x^2-4) >1

{2x^2-5}/(x^2-4) > 0

( 2x^2-3)/(x-2)(x+2) > 0

(x^2-5/2)/ (x+2)(x-2) > 0

f(x) ={x+sqrt(3/2)}(x-sqrt(3/2)} > 0

The critical points are -2, -sqrt(3/2) , sqrt(3/2), 2

All factors are positve when x > 2. So f(x) > 0

f(x) is negative when sqrt(5/2) < x< 2 as numerator is postive and denominator is -ve.

For -sqrt5/2 < x < sqrt(5/2), both Numearator and denominator -ve. So f(x) >0

For -2 < x < -sqrt(5/2) , Numeraror positive, but denominator -ve. so f(x) is negative.

For x< -2, both numerator and denominator are -ve and so f(x) >0.

f(x) = 0 when x= sqrt(5/2) or x = -sqrt(5/2)

f(x) is undefined when x =2 or x=-2 as the denominator ibecomes zero

We'll add +1 both sides E(x)>0.

( x^2 - 1 ) / ( x^2 - 4) +1>0

We'll multiply 1 by the denominator ( x^2 - 4)

(x^2 - 1 + x^2 - 4)/(x^2-4)>0

(2*x^2-5)/(x^2-4)>0

The numerator: f1(x)=2*x^2-5

The denominator f2(x)=x^2-4

We'll discuss the sign of the numerator.

For this reason, we'll calculate the solutions of the equation f1(x)=0

2*x^2-5=0 => 2*x^2=5 => x^2=5/2

x1= sqrt (5/2) and x2=-sqrt (5/2)

For f1(x), with a=2>0, between it's roots,we'll have the opposed sign to "a" sign, so the values of f1(x) are negative and outside the roots, the values are positive.

f1(x)>0 for x belongs to (-inf,-sqrt(5/2))U(sqrt (5/2),inf)

f2(x)<0 for x belongs to (-sqrt(5/2),sqrt(5/2))

We'll discuss the sign of the denominator f2(x)=x^2-4

f2(x)= (x-2)(x+2)

(x-2)(x+2)=0

x1=2 and x2=-2

For f2(x), with a=1>0, between it's roots,we'll have the opposed sign to "a" sign, so the values of f2(x) are negative and outside the roots, the values are positive.

f2(x)>0 for x belongs to (-inf,-2)U(2,inf)

f2(x)<0 for x belongs to (-2,2)

**The intervals where E(x)>0 are:**

**(-inf,-2) U (-sqrt(5/2),sqrt(5/2)) U (2,inf)**