Integration by partial fractions.

Econometricia
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1.\int2x^2+x+9/(9x+1)(x^2+9) dx

2. (A/9x+1) + [(Bx + C ) / (x^2 + 9)]

I get the worst numbers when I solve the system. The question is from an old exam and calculators are not allowed. Am I doing something wrong or is there another way to integrate this?
 
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Are you sure of the numbers you are getting? IOW, did you check to make sure that your two separate fractions add up to the larger one?
 
Yes, I used a solver to check my work because it did not seem right. I am fairly sure the fractions are broken up properly as well. =(
 
Yes, your decomposition is fine. Out of curiosity, what did you get for A, B, and C?
 
a=37/41 b=5/41 c = 7/41 I mean its not that outrageous I guess. I guess I just assume I do things wrong lol.
 
Just to make sure, the integrand is

\frac{2x^2+x+9}{(9x+1)(x^2+9)}

and not

2x^2+x+\frac{9}{(9x+1)(x^2+9)}

right? In either case, your values for A, B, and C aren't correct. If the first version of the integrand is the right one, the coefficients come out even less pretty.
 
Yes the first version is correct. -.- Do you have any suggestions?
 
your partial fractions expansion is:

<br /> \frac{A}{9 x + 1} + \frac{B \, x + C}{x^{2} + 9}<br />

Multiplying both sides by (9 x + 1)(x^{2} + 9) and comparing similar terms, you will get 3 equations for 3 unknowns.
 
I get the following system:

<br /> \left\{<br /> \begin{array}{rcl}<br /> 9 A + C &amp; = &amp; 9 \\<br /> <br /> B + 9 C &amp; = &amp; 1 \\<br /> <br /> A + 9 B &amp; = &amp; 2<br /> \end{array} \right.<br />
 
  • #10
After you multiply by the denominator, I'd set x=-1/9, which allows you to solve for A, then use the equations Dickfore got to solve for B and C.
 
  • #11
Yeah... It must be some type of cruel joke.
 
  • #12
The denominators for the solutions are equal to the number of days in a year.
 
  • #13
361/365 41/365 36/365
It was on an old exam lol. I hope there's nothing like that on our exam =O.
 
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