# Integration by partial fractions.

Econometricia
1.$$\int$$2x^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?

Mentor
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?

Econometricia
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. =(

Mentor
Yes, your decomposition is fine. Out of curiosity, what did you get for A, B, and C?

Econometricia
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.

Staff Emeritus
Homework Helper
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.

Econometricia
Yes the first version is correct. -.- Do you have any suggestions?

Dickfore

$$\frac{A}{9 x + 1} + \frac{B \, x + C}{x^{2} + 9}$$

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

Dickfore
I get the following system:

$$\left\{ \begin{array}{rcl} 9 A + C & = & 9 \\ B + 9 C & = & 1 \\ A + 9 B & = & 2 \end{array} \right.$$

Staff Emeritus
Homework Helper
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.

Econometricia
Yeah.... It must be some type of cruel joke.

Dickfore
The denominators for the solutions are equal to the number of days in a year.

Econometricia
361/365 41/365 36/365
It was on an old exam lol. I hope there's nothing like that on our exam =O.