# Integration by partial fractions.

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?

Mark44
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?

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. =(

Mark44
Mentor
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.

vela
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.

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

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

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

I get the following system:

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

vela
Staff Emeritus
Homework Helper