Homework Help: Partial Fractions Help - Calc II Integration

1. Sep 20, 2008

demersal

1. The problem statement, all variables and given/known data
$$\int$$1/(s$$^{2}$$(s-1)$$^{2}$$) ds

2. Relevant equations
Partial Fractions

3. The attempt at a solution
= $$\frac{A}{s^{2}}$$+$$\frac{B}{s-1}$$+$$\frac{C}{(s-1)^{2}}$$

Setting numerators equal to each other:
1 = A(s-1)(s-1)$$^{2}$$ + Bs$$^{2}$$(s-1)$$^{2}$$+Cs$$^{2}$$(s-1)
1=(As-A)(s$$^{2}$$-2s+1)+Bs$$^{2}$$(s$$^{2}$$-2s+1)+Cs$$^{3}$$+Cs$$^{2}$$
1=As$$^{3}$$-2As$$^{2}$$+As-As$$^{2}$$+2As-A+Bs$$^{4}$$-2Bs$$^{3}$$+Bs$$^{2}$$+Cs$$^{3}$$-Cs$$^{3}$$+Cs$$^{2}$$
1=Bs$$^{4}$$+(A-2B+C)s$$^{3}$$+(-3A+B-C)s$$^{2}$$+3As-A

Using coefficients to solve for variables (here's where something is up!)
s$$^{4}$$: 0=B
s$$^{3}$$: 0=A-2B+C
s$$^{2}$$: 0=-3A+B-C
s: 0=3A
#: 1=-A

They don't add up! Am I doing something wrong?
Thank you for everyone's time and help!

2. Sep 20, 2008

Dick

Don't you need a D/s term as well? If you can't solve it you are missing a variable. (s+1)^2 is a repeated factor and you have two variables for it. Why not for s^2 as well?

3. Sep 20, 2008

demersal

Ohh okay! I'll give it a try! :)

EDIT: Adding a D/s term only seemed to make the situation worse. Unless I made a slight error, there is now no term A, B, C, or D that stands unattached to an "s" and thus there is no variable to equate to the constant.

Ideas??

Last edited: Sep 20, 2008
4. Sep 21, 2008

bigevil

I'm not sure if you really have to do it by partial fractions, but a simpler way to do this problem is by trigonometric substitution.

5. Sep 21, 2008

HallsofIvy

$$\frac{1}{s^2(s-1)^2}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{s-1}+ \frac{D}{(s-1)^2}$$
$$1= As(s-1)^2+ B(s-1)^2+ Cs^2(s-1)+ Ds^2$$
Taking s= 0, B= 1.
Taking s= 1, D= 1.
Taking s= -1, 1= -4A+ 4B- 2C+ D= -4A+ 4- 2C+ 1 so 4A+ 2C= 4.
Taking s= 2, 1= 2A+ B+ 4C+ 4D= 2A+ 1+ 4C+ 4 so 2A+ 4C= -3

6. Sep 21, 2008

demersal

Thank you so much, Hall of Ivy! I had a feeling I was doing way too much work. Now I understand :)