http://i142.photobucket.com/albums/r118/xcelxp/mq1.jpg

Ive been able to complete a couple of problems. I was hoping someone could help me with this one. Thanks.

Printable View

- Apr 20th 2007, 09:43 AMxcelxpIntegration by partial fractions.
http://i142.photobucket.com/albums/r118/xcelxp/mq1.jpg

Ive been able to complete a couple of problems. I was hoping someone could help me with this one. Thanks. - Apr 20th 2007, 10:16 AMecMathGeek
First, I would use polynomial division to simplify this fraction:

4x^2 - 4 | 3x^3 - 0x^2 - 3x + 4

= 3/4*x + 4/(4x^2 - 4)

= 3/4*x + 1/(x^2 - 1)

So now we have:

INT (3/4*x + 1/(x^2 - 1)) dx

= INT 3/4*x dx + INT 1/(x^2 - 1) dx

= 3/8*x^2 + INT 1/(x^2 - 1) dx

Use Partial fractions on the remaining integration:

INT 1/(x^2 - 1) dx

Factor the denominator:

x^2 - 1 = (x - 1)(x + 1)

Now set up your seperate fractions:

1/(x^2 - 1) = A/(x - 1) + B/(x + 1)

Multiply out the denominator (x^2 - 1) to both sides:

1 = A(x + 1) + B(x - 1) = Ax + A + Bx - B

Set terms equal on both sides (x's with x's and constants with constants)

(A + B)x = 0x --> A + B = 0 --> A = -B

(A - B) = 1 --> -B - B = 1 --> B = -1/2, A = 1/2

So the integration becomes:

INT 1/2*1/(x - 1) - 1/2*1/(x + 1) dx = 1/2ln|x - 1| - 1/2ln|x + 1|

The final answer is:

3/8*x^2 + 1/2ln|x - 1| - 1/2ln|x + 1|

Edited: My original solution had 2 as the coefficients of the ln functions when they should be 1/2. - Apr 20th 2007, 01:56 PMKrizalid
- Apr 20th 2007, 02:18 PMecMathGeek
Thank you for pointing out the constant, but are you sure about the (1/2) coefficients? I can go back through my work and see if I missed something, but I'm fairly confident they are 2's and not 1/2's.

You're right. I had a typo in my work that I carried through by accident. I might be off today or something. This isn't my first typo. - Apr 20th 2007, 02:20 PMJhevon
i agree with krizalid, i think the coefficients are (1/2)'s. let me double check

i believe here is where you made an error. you have the numerator as 4 here, when you should have it as 1 according to your method

I have days like that too--and worst days. not only do i make typos on such days, but i misread the question in the first place.

so i try to solve the wrong question AND i do it incorrectly...luckily you guys are always there to back me up:D - Apr 21st 2007, 04:33 AMSoroban
Hello, xcelxp!

I was taught a different procedure for Partial Fractions.

. . Try it . . . I think you'll like it.

. . . . . . . .3x³ - 3x + 4 . . . . . . . . 1

We have: .-------------- .= .¾x + -------

. . . . . . . . .4x² - 4 . . . . . . . . . .x² - 1

Now we decompose that fraction:

. . . . . . 1 . . . . . . . . .A . . . . B

. . ---------------- .= .------ + ------

. . (x - 1)(x + 1) . . . x - 1 . .x + 1

Clear denominators: .1 .= .A(x + 1) + B(x - 1)

.Let x = 1: .1 .= .A(2) + B(0) . . → . A = ½

Let x = -1: .1 .= .A(0) + B(-2) . → . B = -½

. . . . . . . . . . . . . . . . . . . . . . . . .½ . . . ½

Hence, we must integrate: .¾x + ------ - ------

. . . . . . . . . . . . . . - . . . . . . . . .x - 1 - x + 1