# calculus

posted by on .

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!

• calculus - ,

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

• calculus - ,

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...