Write out the form of the partial fraction decomposition of the function appearing in the integral:
integral (6x-58)/(x^2+2x-63)
Determine the numerical values of the coefficients, A and B, where A <= B
A=?
B=?
Thank you guys so much!
=)
I will split it up for you ...
(6x-58)/( (x+9)(x-7) )
let
A/(x+9) + B/(x-7) = (6x-58)/( (x+9)(x-7) )
A(x-7) + B(x+9) = 6x - 58
let x = 7
0 + 16B = -16
B = -1
let x = -9
-16A + 0 = -112
A = -7
so ....
(6x-58)/( (x+9)(x-7) ) = -7/(x+9) - 1/(x-7)
now it integrates easily using ln
take it from here
Thanks! Got it!
To perform partial fraction decomposition of the given function,
1. Factorize the denominator:
x^2 + 2x - 63 = (x + 9)(x - 7)
2. Write out the general form of the partial fraction decomposition:
(6x - 58)/(x^2 + 2x - 63) = A/(x + 9) + B/(x - 7)
3. Multiply both sides of the equation by the denominator to simplify the numerator:
6x - 58 = A(x - 7) + B(x + 9)
4. Since this equation should hold for all values of x, we can choose specific values for x that will cancel one variable term at a time:
For x = 7, we get: 6(7) - 58 = B(7 + 9)
42 - 58 = 16B
-16 = 16B
B = -1
For x = -9, we get: 6(-9) - 58 = A(-9 - 7)
-54 - 58 = -16A
-112 = -16A
A = 7
Thus, the partial fraction decomposition is:
(6x - 58)/(x^2 + 2x - 63) = 7/(x + 9) - 1/(x - 7)
Therefore, A = 7 and B = -1.
To find the partial fraction decomposition of the function (6x - 58)/(x^2 + 2x - 63), we need to factor the denominator.
Step 1: Factor the denominator
The denominator x^2 + 2x - 63 can be factored as (x - 7)(x + 9).
Step 2: Write the partial fraction decomposition
Now, we can write the given function as the sum of two fractions:
(6x - 58)/(x^2 + 2x - 63) = A/(x - 7) + B/(x + 9)
Step 3: Find the numerical values of A and B
To find the values of A and B, we need to clear the fraction by multiplying each fraction by its respective denominator:
(6x - 58) = A(x + 9) + B(x - 7)
Simplify this equation to solve for A and B.
6x - 58 = Ax + 9A + Bx - 7B
Now, group the x terms and the constant terms:
6x + Bx + Ax = 9A - 7B + 58
Combine like terms:
(6 + B)x + Ax = 9A - 7B + 58
Since the left-hand side of this equation is in terms of x and the right-hand side is constant, both sides must be equal to each other. This implies that the coefficients of the x terms and the constant terms must be equal.
Equating the coefficients of x:
6 + B = A
Equating the constant terms:
9A - 7B + 58 = 0
We now have a system of two equations with two unknowns. Solving these equations will give us the values of A and B.
From the first equation, we can isolate B:
B = A - 6
Substitute this value of B into the second equation:
9A - 7(A - 6) + 58 = 0
Simplify this equation and solve for A:
9A - 7A + 42 + 58 = 0
2A + 100 = 0
2A = -100
A = -50
Now substitute this value of A back into the equation for B to find B:
B = A - 6
B = -50 - 6
B = -56
Therefore, the values of A and B are:
A = -50
B = -56
Thus, the partial fraction decomposition of the given function is:
(6x - 58)/(x^2 + 2x - 63) = -50/(x - 7) - 56/(x + 9)
And A <= B as required.