I'm having trouble trying to solve for the partial fraction decomposition in order to find the integral.
∫ 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!
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
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.
To find the values of B, C, D, and E in the partial fraction decomposition, you can use a technique called the method of partial fractions. Here's how you can proceed:
1. Start with the expression for x:
x = B(x^2 + a^2)(x - a) + C(x + a)(x^2 + a^2) + (Dx + E)(x + a)(x - a)
2. Expand and simplify the right side of the equation:
x = B(x^3 - ax^2 + a^2x - a^3) + C(x^3 + ax^2 + a^2x + a^3) + (Dx + E)(x^2 - a^2)
Simplify further by combining like terms:
x = (B + C)x^3 + (a^2B + a^2C + Dx + E)x^2 + (aB + aC - a^3 + Dx - Ex) + a^2B + a^2C)
3. Equate the coefficients of the corresponding powers of x on both sides of the equation. This will give you a system of equations to solve for the unknowns B, C, D, and E.
Coefficient of x^3:
B + C = 0
Coefficient of x^2:
a^2B + a^2C + D = 1
Coefficient of x:
aB + aC - a^3 + D - E = 0
Constant term (coefficient of x^0):
a^2B + a^2C = 0
4. Solve the system of equations to find the values of B, C, D, and E.
From the constant term equation, we have a^2B + a^2C = 0, which implies B = -C.
Substituting B = -C into the coefficient of x^2 equation gives:
a^2(-C) + a^2C + D = 1
D = 1
Substituting B = -C into the coefficient of x equation gives:
a(-C) + aC - a^3 + 1 - E = 0
-a^3 + 1 - E = 0
E = -a^3 + 1
Since B = -C, we can represent B as -P and C as P for convenience, where P is a constant.
Finally, we have:
B = -P
C = P
D = 1
E = -a^3 + 1
5. Substitute these values back into the partial fraction decomposition equation:
x = -P(x^2 + a^2)(x - a) + P(x + a)(x^2 + a^2) + (x + a)(x - a)
6. Now, you can integrate the equation:
∫ x dx = -P∫(x^2 + a^2)(x - a) dx + P∫(x + a)(x^2 + a^2) dx + ∫(x + a)(x - a) dx
7. Simplify and integrate each term to find the final solution.
This method allows you to decompose a rational function into simpler fractions, making it easier to integrate.