Partial fraction decomposition

In summary, Daniel is trying to figure out how to solve an integral using partial fraction decomposition, but makes a mistake with the coefficient of x^2.
  • #1
RadiationX
256
0
Use partial fraction decompostion to find:
[tex]\int_{a}^{b} \frac{2x-1}{x^2(3x+1)(x^2 + 1)}[/tex]

is this partial fraction set up correct?

[tex]\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = 2x - 1[/tex]

If this is correct i can solve the integral.
 
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  • #2
Almost. It's

[tex]\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = \frac{2x - 1}{x^2(3x+1)(x^2+1)}[/tex]

I'm sure you meant that anyways, though...
 
  • #3
yeah that's what i meant. thanks
 
  • #4
this problem is harder than i thought are the following answers correct:

A = 27/10
B = -1
C = 1/10
D = -7/10
E = 1
 
  • #5
I get

[tex]A = -\ \frac{27}{2}, \; B = C = - \ \frac{1}{2}, \; D = 5, \; E = -1.[/tex]
 
  • #6
i don't see how Data. i believe you but i can't reproduce those numbers.
how did you do it?
 
  • #7
my first number A = 27/10 is wrong so that makes every thing that follows wrong.
 
  • #8
OMG! I'm using 2x +1 instead of the correct 2x-1
 
  • #9
First multiply each side through by all three factors in the denominator on the right to get

[tex]Ax^2(x^2+1) \ + \ (Bx+C)(3x+1)x^2 \ + \ (Dx+E)(x^2+1)(3x+1) = 2x-1[/tex]

now sub [itex]x=0[/itex] to immediately give [itex]E=-1[/itex]. Then sub [itex]x=-1/3[/itex] to give

[tex]A\left(-\frac{1}{3}\right)^2\left(\left(-\frac{1}{3}\right)^2+1\right) = -\frac{5}{3} \Longrightarrow A = -\frac{27}{2}[/tex]

After that, look at the coefficient of [itex]x[/itex] on both sides. On the left it is [itex]D+3E[/itex] and on the right [itex]2[/itex], so

[tex]D+3E = D - 3 = 2 \Longrightarrow D = 5.[/tex]

Now look at the coefficient of [itex]x^2[/itex] on both sides. On the left it is [itex]A + C + 3D + E[/itex] and on the right [itex]0[/itex] so

[tex] A + C + 3D + E = -\frac{27}{2} + C + 15 - 1 = 0 \Longrightarrow C = -\frac{1}{2}.[/tex]

Finally look at the coefficient of [itex]x^3[/itex] on both sides. On the left it is [itex]B+3C+D+3E[/itex] and on the right it is [itex]0[/itex] so we get

[tex]B + 3C + D + 3E = B - \frac{3}{2} + 5 - 3 = 0 \Longrightarrow B = -\frac{1}{2}[/tex]

so overall we have found

[tex] E = -1, \; A = -\frac{27}{2}, \; D = 5, \; C = B = -\frac{1}{2}[/tex]

as I said :smile:
 
  • #10
I've been making dumb mistakes all day. thanks for taking the time to latex your replies , and thanks for all the help.
 
  • #11
No problem :smile:

I latex everything that I can anyways, better to get used to it now!
 
  • #12
[tex]\frac{2x-1}{x^2\left( 3x+1\right) \left( x^2+1\right) }=\allowbreak -\frac 1{x^2}+\frac 5x-\frac{27}{2\left( 3x+1\right) }-\frac 12\frac{x+1}{x^2+1}[/tex]

Daniel.

P.S.Just trying my Maple :rolleyes: :-p ...
 

Related to Partial fraction decomposition

What is partial fraction decomposition?

Partial fraction decomposition is a mathematical technique used to break down a rational expression, which is a fraction where the numerator and denominator are both polynomials, into simpler components. This is done by expressing the rational expression as a sum of smaller fractions with a single polynomial in the denominator.

Why is partial fraction decomposition useful?

Partial fraction decomposition is useful in solving integrals, differential equations, and other mathematical problems involving rational expressions. It allows us to simplify complex expressions and make them easier to work with.

How is partial fraction decomposition done?

The steps for performing partial fraction decomposition include: 1. Factor the denominator of the rational expression. 2. Write the rational expression as a sum of fractions with the factors of the denominator as the individual denominators. 3. Solve for the unknown coefficients in each fraction by equating the original expression to the partial fraction decomposition. 4. Combine like terms and simplify the resulting expression.

What types of rational expressions can be decomposed using partial fraction decomposition?

Partial fraction decomposition can be used for proper rational expressions, which are expressions where the degree of the numerator is less than the degree of the denominator. It can also be used for improper rational expressions, but additional steps may be required.

Are there any limitations or restrictions to using partial fraction decomposition?

Yes, there are some limitations and restrictions to using partial fraction decomposition. The denominator of the rational expression must be a polynomial, and it must factor into linear and irreducible quadratic factors. The coefficients in the partial fraction decomposition must also be real numbers.

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