Need help finding some unknown constants

  • Thread starter 1MileCrash
  • Start date
  • #1
1MileCrash
1,339
41

Homework Statement



I have an integral that I'm trying to split into partial fractions, and I've gotten to an equation but I'm not sure how to solve this one.

Homework Equations





The Attempt at a Solution



93x+2 = Ax^2 + Bx + A - 4B

I have no idea, because I usually don't have a x^2 term to deal with. I assumed that A should be 0 because of this, but that requires B to equal 93, and for A - 4B to equal 2. So, that's not the right direction.
 

Answers and Replies

  • #2
flyingpig
2,580
1
Yes A = 0

and A - 4B = 2

You know what A is...
 
  • #3
1MileCrash
1,339
41
How does that work? That means B is - 1/2, and

93x + 2 = -1/2x + 2

Is not true.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,693
1,273
Yes A = 0

and A - 4B = 2

You know what A is...
And this requires B = -1/2 , but that contradicts B = 93.

So, there appears to be something wrong with the way you obtained your equation: 93x+2 = Ax2 + Bx + A - 4B.

Show us how you came up with that so we can help you.
 
  • #5
1MileCrash
1,339
41
Okay

[itex]\int \frac{93x+2}{x^{3}-4x^{2}+x-4}dx[/itex]

I factored this by finding the factor of 4 that yields a 0 when plugged into the denominator, which is 4 itself. So I concluded one term is (x-4), and found the other through polynomial long division.

[itex]\int \frac{93x+2}{(x-4)(x^{2}+1)}dx[/itex]

So I then attempted to split this into partial fractions.


[itex]\int \frac{A}{(x-4)}+ \frac{B}{(x^{2}+1)}dx[/itex]

So then

93x + 2 = Ax^2 + A + Bx - 4B

What's wrong with the equation?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,693
1,273
That needs to be [itex]\displaystyle \frac{A}{x-4}+ \frac{Bx + C}{x^{2}+1}[/itex]
 
  • #7
1MileCrash
1,339
41
Oh, ok. Why is that? Because I have a variable in my numerator?
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,693
1,273
The denominator x2 + 1 is quadratic in x, so the most general numerator is linear in x.
 
  • #9
1MileCrash
1,339
41
When working it I get:

A = 23.25
B = - A
C = 5.3125

Agree?
 
  • #10
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,693
1,273
No, although, B = -A is correct.

Did you get

93x + 2 = A(x2 + 1) +(Bx +C)(x - 4) ?

Hint: Set x = 4 to find A.
 
  • #11
1MileCrash
1,339
41
I get

x^2(A+B) + x(C-4B) - 4C + A = 93x + 2

From there, I said that A+B = 0, C-4B = 93, and -4C + A = 2 and I am getting it wrong every time.
 

Suggested for: Need help finding some unknown constants

Replies
9
Views
730
  • Last Post
Replies
14
Views
777
Replies
4
Views
299
  • Last Post
Replies
15
Views
784
Replies
3
Views
498
Replies
5
Views
780
Replies
7
Views
722
Replies
15
Views
746
Replies
2
Views
441
Top