Partial Fraction Integration

1. May 6, 2007

tanky322

1. The problem statement, all variables and given/known data

integral((3X+4)/((x^2+4)(3-x))

2. Relevant equations

3. The attempt at a solution
I know I should be using partial fractions for this problem but the x^2 in the denom. screws me up. I think the partial fraction should be:

(Ax+B)/(x^2+4) + C/(3-x)

Then:

(Ax+B)(3-x)+C(x^2+4)

Then:

(x^2+4)(3-x)=-Bx-Ax^2+3B+3Ax+Cx^2+C4

So now Im completely lost assuming im even on the right track.

Thanks Alot!!!

2. May 6, 2007

Office_Shredder

Staff Emeritus
This makes no sense. Your first steps aren't actual equations, which is confusing you.

(Ax+B)/(x^2+4) + C/(3-x) = (3x+4)/(x^2+4)(3-x)

Multiplying by the two denomiator parts gives

(Ax+B)(3-x) + c(x^2+4) = 3x+4

This is for all x, so plug in clever values (like x=3), to solve for a, b and c

3. May 6, 2007

cristo

Staff Emeritus
I'm not sure what you're doing! You have $$\frac{3x+4}{(x^2+4)(3-x)}=\frac{Ax+B}{x^2+4}+\frac{C}{3-x}$$. Now multiply through by the denominator of the LHS to yield $$3x+4=(Ax+B)(3-x)+C(x^2+4)$$. Now you need to find three equations to obtain A B and C.

4. May 6, 2007

tanky322

Ok it makes more sense now, this partial fraction stuff drives me nuts!:yuck:

Thanks guys!

5. May 7, 2007

tanky322

Ok so I just worked on this problem again.... It was driving me nuts! This is what I came out with,
A=1
B=0
C=1

So for a final answer I came up with,
(Ln(x^2+4))/2+Ln(3-x)+ C

Is that right??

Thanks,

Andy

6. May 7, 2007

Curious3141

Partial fraction decomposition right. Integration wrong. The second term should have a negative sign. The answer should be $$\frac{1}{2}\ln{(x^2+4)} - \ln{|3-x|} + C$$. Don't forget the modulus sign on the second term (it's optional for the first since the square expression is nonnegative for real x).