Express
x^2 - 3x
x^2 - 1
As partial fractions
To express the given expression as partial fractions, we need to first factorize the denominators of the two fractions.
The denominator of the first fraction, x^2 - 1, can be factorized as (x - 1)(x + 1).
The denominator of the second fraction, x^2 - 3x, can be factorized as x(x - 3).
Now, let's express the given expression as partial fractions:
(x^2 - 3x) / (x^2 - 1) = A/(x - 1) + B/(x + 1) + C/(x(x - 3))
To determine the values of A, B, and C, we need to add the fractions on the right side together and equate the numerators on both sides of the equation:
x^2 - 3x = A(x + 1)(x(x - 3)) + B(x - 1)(x(x - 3)) + C(x - 1)(x + 1)
Expanding and simplifying the equation:
x^2 - 3x = A(x^3 - 2x^2 - 3x + 3) + B(x^3 - 4x^2 + 3x - 3) + C(x^2 - x - x + 1)
Now, let's collect like terms:
x^2 - 3x = (A + B)x^3 + (-2A - 4B + C)x^2 + (-3A + 3B - C)x + (3A - 3B)
Since both sides of the equation are polynomials, the coefficients of the corresponding powers of x must be equal. Equate the coefficients of each corresponding power of x:
For x^3: A + B = 0 (coefficient of x^3 on the left is 0)
For x^2: -2A - 4B + C = 1 (coefficient of x^2 on the left is 1)
For x: -3A + 3B - C = -3 (coefficient of x on the left is -3)
For constant term: 3A - 3B = 0 (constant term on the left is 0)
Now, we have a system of linear equations with A, B, and C as unknowns. Solving this system of equations will give us the values of A, B, and C, which can then be substituted back into the partial fractions expression.