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?

 

[Mark44]

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. =(

 

[Mark44]

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.

 

[vela]

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]

your partial fractions expansion is:

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

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:

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

 

[vela]

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.