Prove that (x-1)*f(x)>0 if f(x)=(x^4+x)/x(x^2-1)

We need to factorize the numerator of the
function:

f(x) = x(x^3 + 1)/x(x^2 -
1)

We'll simplify:

f(x) = (x^3
+ 1)/(x^2 - 1)

The sum of cubes form numerator returns the
product:

x^3+1 = (x+1)(x^2 - x +
1)

The difference of squares form numerator returns the
product:

x^2 - 1 =
(x-1)(x+1)

f(x) = (x+1)(x^2 - x +
1)/(x-1)(x+1)

We'll
simplify:

f(x) = (x^2 - x +
1)/(x-1)

Now, we have to verify if
(x-1)*f(x)>0

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

We'll simplify:

(x^2 - x +
1) > 0

We'll verify if the parabola given by the
quadratic expression is above x axis.

For this reason,
we'll check the value of it's discriminant:

delta = b^2 -
4ac

a = 1, b = -1 , c =
1

delta = 1 - 4 = -3 <
0

Since delta is negative and a = 1>0, the parabola
given by the quadratic expression is above x
axis.

Therefore, the inequality
(x-1)*f(x)>0 is verified.