To find the probability that X^2 lies between 1/3 and 1, we need to integrate the density function f(x) over the given interval.

Given:
Density function f(x) = c/(x^2 - 1)

Step 1: Determining the constant c
To find the constant c, we need to ensure that the total area under the density function is equal to 1. The total area under any probability density function should always be equal to 1.

∫f(x)dx = 1

∫c/(x^2 - 1)dx = 1

To integrate this, we can use the method of partial fractions. Factoring the denominator, we get:

∫c/((x-1)(x+1))dx = 1

Using partial fractions, we can express the integrand as:

c/((x-1)(x+1)) = A/(x-1) + B/(x+1)

Multiplying both sides by (x-1)(x+1), we get:

c = A(x+1) + B(x-1)

Expanding and equating coefficients, we have:

A + B = 0 (coefficient of x^1 term)
A - B = c (constant term)

Solving these equations, we get A = B = c/2.

Therefore, c = A = B = 2.

Step 2: Integrating f(x) over the given interval
Now that we have determined the value of c, we can integrate f(x) over the interval [1/3, 1] to find the probability.

P(1/3 ≤ X^2 ≤ 1) = ∫[1/3, 1] f(x)dx

Substituting the value of c, we have:

P(1/3 ≤ X^2 ≤ 1) = ∫[1/3, 1] (2/(x^2 - 1))dx

We can now integrate this expression to find the probability.

∫(2/(x^2 - 1))dx = 2∫(1/(x-1)(x+1))dx

Using partial fractions again, we can express the integrand as:

1/(x-1)(x+1) = A/(x-1) + B/(x+1)

Multiplying both sides by (x-1)(x+1), we get:

1 = A(x+1) + B(x-1)

Expanding and equating coefficients, we have:

A + B = 0 (coefficient of x^1 term)
A - B = 1 (constant term)

Solving these equations, we get A = 1/2 and B = -1/2.

Therefore, the integral becomes:

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

Simplifying, we get:

2(1/2 ln|x-1| - 1/2 ln|x+1|) evaluated from 1/3 to 1

Substituting the limits of integration, we have:

2(1/2 ln|1-1| - 1/2 ln|1+1|) - 2(