**Integration of (x^2 - 1)/(x^4 + x^2 + 1) dx from 0 to 1**

To find the value of the given integral, we first need to simplify the integrand. Let's rewrite the integrand as a partial fraction:

x^2 - 1 = (x^4 + x^2 + 1) * A + B * (x^2 + 1)

Expanding the right side:

x^2 - 1 = Ax^4 + Ax^2 + A + Bx^2 + B

Matching the coefficients of like powers of x:

1. Coefficient of x^4: A = 0
2. Coefficient of x^2: A + B = 1
3. Constant term: A = -1

From equation 1, A = 0, so substituting A = 0 into equation 2:

0 + B = 1
B = 1

Now we can rewrite the integrand:

(x^2 - 1)/(x^4 + x^2 + 1) = (1 * (x^2 + 1))/(x^4 + x^2 + 1)

Let's break down the integral into two parts:

∫[(1 * (x^2 + 1))/(x^4 + x^2 + 1)] dx = ∫(1/x^2) dx + ∫(1/(x^2 + 1)) dx

**Evaluating the first integral:**

∫(1/x^2) dx = ∫x^(-2) dx
= x^(-1)/(-1) + C
= -1/x + C

**Evaluating the second integral:**

∫(1/(x^2 + 1)) dx = tan^(-1)(x) + C

Now we can evaluate the definite integral from 0 to 1:

∫[(1 * (x^2 + 1))/(x^4 + x^2 + 1)] dx from 0 to 1
= [-1/x + tan^(-1)(x)] from 0 to 1
= (-1/1 + tan^(-1)(1)) - (-1/0 + tan^(-1)(0))
= (-1 + π/4) - (undefined + 0)
= -1 + π/4

Thus, the value of the given integral is -1 + π/4.

**Proving the value is π/2√3:**

-1 + π/4 = -4/4 + π/4
= (-4 + π)/4
= π/4 - 1

Now, we need to show that π/4 - 1 is equal to π/2√3.

To do this, we can rationalize the denominator of π/2√3:

π/2√3 = (π/2√3) * (√3/√3)
= (π√3)/(2√3)
= π/(2√3)

By comparing the two expressions, we can see that π/(2√3) is equal to π/4 - 1.

Therefore, we have proven that the value of the given integral is