Question

Stuck on a partial fraction integral problem:

Answer 1

Firstly:
\[\int\limits_{}^{}(x ^{2}+11)/((x-3)(x ^{2}+1)) = \int\limits_{}^{}A/(x-3) + \int\limits_{}^{}(Bx+C)/(x ^{2}+1)\]
Find variables:
Mulitply by LCD after setting up:\[x ^{2}+11=A(x ^{2}+1)+Bx+C(x-3)\]
Distribute on right:
\[= Ax ^{2}+A+Bx ^{2}-3Bx+Cx-3C\]
Combine like terms:
\[(A+B)x ^{2}+(-3B+C)x+(-3C+A)x ^{0}\]
Equate coefficients of like terms on opposite sides of equation:
\[x ^{0}: 11 = -3C+A\]
\[x: 0=-3B+C\]
\[x ^{2}:1=A+B\]

Answer 2

Now from equating the coefficients of x I know I have C = 3B which I can then plug in for c in the top?

Answer 3

The x^0 equation

Answer 4

But I am not sure if that is allowed or if I am solving the system of equation it makes after plugging in

Answer 5

Don't know why I can't see the latex here, so pls attach the question

Answer 6

Dang it, can anyone see it?

Answer 7

http://imgur.com/ZXOdQ

Answer 8

There's a screen shot

Answer 9

So the systems of equation I make is:
11=-9B+A <<(new version of X^0 equation, since C=3B from x equation)
+9=9B+9A<<(by multiplying x^2 equation by 9)
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So, 19=10A : A = 19/10

Answer 10

And that's where I am at right now I am skeptical I have been correct up till now so I just want to make sure before I proceed.

Answer 11

I got A=2, B=-1, and C=-3

Answer 12

Yeap I also got A=2,B=-1 and C=-3

Answer 13

What am I doing wrong?

Answer 14

Should have been Ax^2+A+Bx+Cx-3C

Answer 15

All I did was distribute out during this step:
A(x2+1)+Bx+C(x−3)

Answer 16

A+B=1
C-3B=0
A-3C=11
let C=3B then A-3C=11 turns into A-3(3B)=11 so A-9B=11, since A+B=1 then B=1-A, so A-9B=11 turns into A-9(1-A)=11, so A-9+9A=11 so 10A=11+9, 10A=20 and A=2

Answer 17

Distributing that doesn't yield: Ax^2+A+Bx+Cx-3C, that is missing the Bx^2 no?

Answer 18

yes it is missing a bx^2

Answer 19

your first post on the decomposition was correct\[Ax2+A+Bx2−3Bx+Cx−3C \]

Answer 20

So my attempt at setting up a system by muliplying by 9 was incorrect?

Answer 21

Oh wait.. it would have worked, I multiplied wrong

Answer 22

haha

Answer 23

Thanks

Answer 24

Cheers

Answer 25

you know how to do the rest?