Express 1/(x+1)(x-1) in partial fractions?
Let 1/(x+1)(x-1) = A/(x+1) + B/(x-1)
= ( A(x-1) + B(x+1) )/((x-1)(x+1) )
or
A(x-1) + B(x+1) = 1
let x = 1, 2B = 1 ----> B = 1/2
let x = -1, -2A=1 -----> A = -1/2
so 1/(x+1)(x-1)
= 1/(2(x-1)) + 1/(2(x+1))
or (1/2))(1/(x-1)) + 1/(x+1) )
last two lines should be ...
= 1/(2(x-1)) - 1/(2(x+1))
or (1/2))(1/(x-1)) - 1/(x+1) )
To express the fraction 1/(x+1)(x-1) in partial fractions, follow these steps:
Step 1: Factor the denominator (x+1)(x-1)
The denominator (x+1)(x-1) can be factored into two linear factors: (x+1) and (x-1).
Step 2: Write the partial fraction decomposition
Since we have two factors in the denominator, the partial fraction decomposition will have two terms.
1/(x+1)(x-1) = A/(x+1) + B/(x-1)
Step 3: Clear the fractions
To clear the fractions, multiply both sides of the equation by the common denominator (x+1)(x-1):
1 = A(x-1) + B(x+1)
Step 4: Solve for the unknowns A and B
Expand the equation and collect like terms:
1 = (A + B)x - A + B
The coefficients of the x term on each side of the equation must be equal, as well as the constants:
A + B = 0 (coefficient of x)
-A + B = 1 (constant term)
Solve this system of equations to find the values of A and B.
From the first equation, A = -B.
Substitute this value into the second equation:
-(-B) + B = 1
2B = 1
B = 1/2
Since A = -B, A = -1/2
So, the partial fraction decomposition is:
1/(x+1)(x-1) = -1/2/(x+1) + 1/2/(x-1)