Integration of Rational Functions by Partial Fractions

[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

[snipez90]

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

[afcwestwarrior]

woops.

[afcwestwarrior]

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

[afcwestwarrior]

this one is strange

[afcwestwarrior]

How about this 3x (1/3x+3)

yea or nah

[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

[salman213]

How about this 3x (1/3x+3)

yea or nah


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

it = (x^2 + 9x)

[afcwestwarrior]

Oright I'll use your technique. Thanks man.

[afcwestwarrior]

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

[salman213]

correct

[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

[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.

[afcwestwarrior]

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

[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..

[afcwestwarrior]

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

[afcwestwarrior]

Thanks a lot man.

[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

[afcwestwarrior]

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

[afcwestwarrior]

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

[afcwestwarrior]

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

[afcwestwarrior]

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

[HallsofIvy]

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