Need help finding some unknown constants

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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.
 
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Yes A = 0

and A - 4B = 2

You know what A is...
 
How does that work? That means B is - 1/2, and

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

Is not true.
 
flyingpig said:
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.
 
Okay

\int \frac{93x+2}{x^{3}-4x^{2}+x-4}dx

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.

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

So I then attempted to split this into partial fractions.


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

So then

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

What's wrong with the equation?
 
That needs to be \displaystyle \frac{A}{x-4}+ \frac{Bx + C}{x^{2}+1}
 
Oh, ok. Why is that? Because I have a variable in my numerator?
 
The denominator x2 + 1 is quadratic in x, so the most general numerator is linear in x.
 
When working it I get:

A = 23.25
B = - A
C = 5.3125

Agree?
 
  • #10
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
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.
 
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