# Partial Fraction Decomposition with Integration

1. Feb 25, 2014

### jdawg

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

∫(2x3-4x-8)/(x2-x)(x2+4) dx

2. Relevant equations

3. The attempt at a solution

∫(2x3-4x-8)/x(x-1)(x2+4) dx
Next I left off the integral sign so I could do the partial fractions:

2x3-4x-8=(A/x)+(B/(x-1))+((Cx+D)/(x2+4))

2x3-4x-8=A(x3-x2+4x-4)+B(x3+4x)+(Cx+D)(x2-x)

2x3-4x-8=x3(A+B+C)-x2(A+C-D)+x(4A+4B-D)-4A

2=A+B+C

0=A+C-D

-4=4A+4B-D

-8=-A(4)
A=2

Did I set this up correctly? I'm not entirely sure how to solve for these variables.

2. Feb 25, 2014

### scurty

Looks good to me so far, besides a few parentheses errors. Remember if a number is in the denominator, it needs to have parentheses around the whole number.

You figured out A, good. Now you are left with three equations and three unknowns. Plug in your value for A into all of them. Now choose any equation and solve for a variable (try to avoid fractions if you can). Now plug this variable into another equation and solve for either variable. Finally, plug that variable into the final equation to solve for one variable. Now repeat the process until all the variables are solved for.

For example, in the two variable case you may have $A+B=3, A-B=1$. Solving for B you get B = 3 - A, then A - (3 - A) = 1, 2A - 3 = 1, 2A = 4, A = 2. Plug it back in to solve for B.

3. Feb 25, 2014

### jdawg

Thanks so much! I figured it out :)

4. Feb 25, 2014

### scurty

Nice job! I tried to make my explanation as clear as possible but it turned out to be verbose; I'm sure it could have been explained easier.

5. Feb 25, 2014

### jdawg

No, it was great! I understood you perfectly :)