# Partial fraction decomposition

1. Apr 3, 2005

Use partial fraction decompostion to find:
$$\int_{a}^{b} \frac{2x-1}{x^2(3x+1)(x^2 + 1)}$$

is this partial fraction set up correct?

$$\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = 2x - 1$$

If this is correct i can solve the integral.

2. Apr 3, 2005

### Data

Almost. It's

$$\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = \frac{2x - 1}{x^2(3x+1)(x^2+1)}$$

I'm sure you meant that anyways, though...

3. Apr 3, 2005

yeah that's what i meant. thanks

4. Apr 3, 2005

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. Apr 3, 2005

### Data

I get

$$A = -\ \frac{27}{2}, \; B = C = - \ \frac{1}{2}, \; D = 5, \; E = -1.$$

6. Apr 3, 2005

i don't see how Data. i believe you but i can't reproduce those numbers.
how did you do it?

7. Apr 3, 2005

my first number A = 27/10 is wrong so that makes every thing that follows wrong.

8. Apr 3, 2005

OMG!!! i'm using 2x +1 instead of the correct 2x-1

9. Apr 3, 2005

### Data

First multiply each side through by all three factors in the denominator on the right to get

$$Ax^2(x^2+1) \ + \ (Bx+C)(3x+1)x^2 \ + \ (Dx+E)(x^2+1)(3x+1) = 2x-1$$

now sub $x=0$ to immediately give $E=-1$. Then sub $x=-1/3$ to give

$$A\left(-\frac{1}{3}\right)^2\left(\left(-\frac{1}{3}\right)^2+1\right) = -\frac{5}{3} \Longrightarrow A = -\frac{27}{2}$$

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

$$D+3E = D - 3 = 2 \Longrightarrow D = 5.$$

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

$$A + C + 3D + E = -\frac{27}{2} + C + 15 - 1 = 0 \Longrightarrow C = -\frac{1}{2}.$$

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

$$B + 3C + D + 3E = B - \frac{3}{2} + 5 - 3 = 0 \Longrightarrow B = -\frac{1}{2}$$

so overall we have found

$$E = -1, \; A = -\frac{27}{2}, \; D = 5, \; C = B = -\frac{1}{2}$$

as I said

10. Apr 3, 2005

I've been making dumb mistakes all day. thanks for taking the time to latex your replies , and thanks for all the help.

11. Apr 3, 2005

### Data

No problem

I latex everything that I can anyways, better to get used to it now!

12. Apr 4, 2005

### dextercioby

$$\frac{2x-1}{x^2\left( 3x+1\right) \left( x^2+1\right) }=\allowbreak -\frac 1{x^2}+\frac 5x-\frac{27}{2\left( 3x+1\right) }-\frac 12\frac{x+1}{x^2+1}$$

Daniel.

P.S.Just trying my Maple :uhh: :tongue2: ...