I am working on this problem and having some trouble. We're supposed to use partial fractions. The problem is: integral of dx / (x^6-1) . I got the values of A and D, but I am having trouble with the others. Help please!
I am not certian of the fractions you are using.
x^6-1 is a difference of two squares
(x^3-1)(x^3+1) Now both of those are the difference (or sum) of two cubes, which can both be factored. So you have four factors in the denominator. Frankly, I doubt if we can do much in ASCII to help you on those fractions. Notice you have a double quadratic in the denominator.
If you want a computer to assist you, put 1/(x^6-1) in the window here.
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=partial_fractions&s3=basic
I did do the sum and difference of the cubes and got these as my partial fractions:
A / (x+1)
Bx + C / (x^2-x+1)
D / (x-1)
Ex + F / (x^2+x+1)
I got the answer for A = -1/6 and D = 1/6.
But I was having trouble with the other variables...
To find the values of A, B, and C in the partial fraction decomposition, we start by factoring the denominator as a product of irreducible factors.
The denominator, x^6 - 1, can be factored using the formula for the difference of squares: a^2 - b^2 = (a - b)(a + b).
The factors of x^6 - 1 can be written as (x^3 - 1)(x^3 + 1).
Now, let's factor each of these factors further:
x^3 - 1 can be factored as (x - 1)(x^2 + x + 1).
x^3 + 1 can be factored as (x + 1)(x^2 - x + 1).
So, the factorization of x^6 - 1 becomes: (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1).
Next, we express the integrand as a partial fraction:
1 / (x^6 - 1) = A / (x - 1) + B / (x^2 + x + 1) + C / (x + 1) + D / (x^2 - x + 1).
To determine the values of A, B, C, and D, we can use the method of equating coefficients. By multiplying both sides of the equation by (x^6 - 1), we can cancel out the denominators on the right side. This results in the following equation:
1 = A(x^2 + x + 1)(x + 1)(x^2 - x + 1) + B(x - 1)(x + 1)(x^2 - x + 1) + C(x - 1)(x^2 + x + 1)(x^2 - x + 1) + D(x - 1)(x^2 + x + 1)(x + 1).
Now, we substitute values for x that will help us determine the coefficients.
Let's start with x = 1. By substituting this value into the equation, we obtain:
1 = 4A.
Therefore, A = 1/4.
Next, let's try x = -1. By substituting this value into the equation, we get:
1 = 4C.
Hence, C = 1/4.
For B and D, we can use the same approach by choosing values for x, but it is more efficient to use algebraic methods. We can compare the coefficients of corresponding powers of x on both sides of the equation.
Comparing the coefficients of x^5 gives:
0 = A + B.
Since A = 1/4, we can solve this equation to find B:
B = -1/4.
Similarly, comparing the coefficients of x^4 gives:
0 = A + C + D.
We know A = 1/4 and C = 1/4, so we can solve this equation to find D:
D = -1/2.
Now, we have the values of A, B, C, and D. We can substitute them back into the partial fraction decomposition:
1 / (x^6 - 1) = 1/4 / (x - 1) - 1/4 / (x^2 + x + 1) + 1/4 / (x + 1) - 1/2 / (x^2 - x + 1).
You can now integrate this expression using the partial fraction decomposition.