I'm having trouble trying to solve for the partial fraction decomposition in order to find the integral.

Question

∫ x / (x^4 - a^4) dx

I'm assuming a is some constant in this case.

So I factored the denominator to this:
(x^4 - a^4) = (x^2 + a^2)(x + a)(x - a)

Which turns each of them into:
x = B/(x+a) + C/(x-a) + Dx+E/(x^2+a^2)

Getting rid of the fractions lead to:

x = B(x^2+a^2)(x-a) + C(x+a)(x^2+a^2) + (Dx+E)(x+a)(x-a)

But I'm lost on exactly what to do next to substitute and find what B, C, and Dx+E is equal to in order to plug back into the integral. Any help is greatly appreciated!

Answer 1

So, you have two polynomials, which are identical. That means all the coefficients on both sides must be the same. That is,

x = B(x^2+a^2)(x-a) + C(x+a)(x^2+a^2) + (Dx+E)(x+a)(x-a)
= B(x^3-ax^2+a^2x-a^3) + C(x^3+ax^2+a^2x+a^3) + D(x^3-a^2x) + E(x^2-a^2)
= (B+C+D)x^3 + (-aB+aC+E)x^2 + (a^2B+a^2C-a^2D)x +(-a^3B+a^3C-a^2E)

equating coefficients, we have

B+C+D = 0
-aB+aC+E = 0
a^2B+a^2C-a^2D = 1
-a^3B+a^3C-a^2E = 0

Now, you just have to solve for B,C,D,E

You may love the algebra, but I moseyed on over to here to see the result:

http://www.wolframalpha.com/input/?i=(inverse+%7B%7B1,1,1,0%7D,%7B-a,a,0,a%7D,%7Ba%5E2,a%5E2,-a%5E2,0%7D,%7B-a%5E3,a%5E3,0,-a%5E2%7D%7D)*%7B%7B0%7D,%7B0%7D,%7B1%7D,%7B0%7D%7D

Answer 2

The equality should read:
x/(x^4-a^4)=B/(x+a) + C/(x-a) + Dx+E/(x^2+a^2)
The following step is correct.

The next step is to expand the right hand side and compare coefficients, so as to set up a system of equations to solve for B,C,D and E.
For example, if you assemble the terms for x^3, you'd get
x^3(D+C+B).
Since the LHS does not have x^3, we conclude that D+C+B=0, that makes 1 equation.
You will end up with one equation for each power of x, (0-3), thus four equations to solve for B,C,D and E.

The other equations are:
(E+aC-aB)x^2=0
(-a^d+a^C+a^2B)x=1
-a^2E-a^3B+a^3C=0

(Check my work)
Solve for B,C,D and E and proceed to integrate.