# Homework Help: Integration of Rational Functions by Partial Fractions

1. Aug 25, 2008

### afcwestwarrior

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

∫ 10/(x-1)(x^2+9)

would i change this into 10/ (x-1) (x+3) (x+3)

then= A/ x-1 + B/ X+3 + C/ x+3

2. Aug 25, 2008

### snipez90

You have a sum of squares. (x^2 + 9) =/= (x+3)(x+3) = x^2 +6x + 9.

3. Aug 25, 2008

### afcwestwarrior

woops.

4. Aug 25, 2008

### afcwestwarrior

do i just leave it like that x^2+9

5. Aug 25, 2008

### afcwestwarrior

this one is strange

6. Aug 25, 2008

### afcwestwarrior

yea or nah

7. Aug 25, 2008

### salman213

if you have a second power the partial fraction you use is Bx+C/D

for ex.

a/(Ax+B)(Cx^2+D)

if (Cx^2+D) cannot be factored then your partial fraction would be

a/(Ax+B)(Cx^2+D) = E/(Ax+B) + Fx+G/(Cx^2+D)

where, a,A,B,C,D,E,F,G are all constants

8. Aug 25, 2008

### salman213

how does 3x(1/3x +3) = (x^2+9)?

it = (x^2 + 9x)

9. Aug 25, 2008

### afcwestwarrior

Oright I'll use your technique. Thanks man.

10. Aug 25, 2008

### afcwestwarrior

So it would be A/x-1 + Bx+C/ x^2+9

11. Aug 25, 2008

### salman213

correct

12. Aug 25, 2008

### afcwestwarrior

ok here's what i did 10= A(x^2+9) + (Bx+C) (x-1)

10= Ax^2 + 9A + Bx^2- Bx + Cx -C
10= (A+B) x^2 + (9A-C) -(B+C)x
ok so A+B=0, 9A-C = 10, B+C =0

is that correct

13. Aug 25, 2008

### salman213

be careful

10= Ax^2 + 9A + Bx^2- Bx + Cx -C

you have

10= (A+B) x^2 + (9A-C) -(B+C)x = (A+B) x^2 + (9A-C) -Bx-Cx = remember that negative is distributive.

14. Aug 25, 2008

### afcwestwarrior

Ok this is where I need help. Bx+Cx,

15. Aug 25, 2008

### salman213

ok i have to go to sleep now lol, hopefully PF wont kill me for telling its easy enough..:(

Just make it (C-B)x

then solve

ur equations are right A+B=0, 9A-C = 10, B+C =0 but the last one should be changed i think you now..

16. Aug 25, 2008

### afcwestwarrior

oh ok, so would it be (B-C)x

17. Aug 25, 2008

### afcwestwarrior

Thanks a lot man.

18. Aug 25, 2008

### salman213

oh shoot another thing, by the looks of it not that im to judge but u I THINK you may need help later in the actual integration so i will give u the formula,

you will need EVENTUALLY need to integrate

A/x^2+B^2
=
(1/a)tan-1(x/a)

just do the integration and you will see one place u will be like how do i integrate

-1/x^2+9
then use the above formula

ok going now, hopefully someone else will help if u get stuck integrating

19. Aug 25, 2008

### afcwestwarrior

I was wondering how the answer in the back of the book had that. Thanks.

20. Aug 25, 2008

### afcwestwarrior

so A=1, C=-1, and B=1

21. Aug 25, 2008

### afcwestwarrior

ok I'm stuck now. I know it would be ∫1/ (x-1) + ????/ x^2+9

22. Aug 25, 2008

### afcwestwarrior

is this correct ∫1/ (x-1) + ∫ x-1/ x^2+9 = ∫1/ (x-1) +∫ x/x^2+9 - ∫1/ x^2 +9

23. Aug 26, 2008

### HallsofIvy

No, it's (C- B) x. Salman213 already told you that.