# Integration of Rational Functions by Partial Fractions

• Jun 19th 2008, 09:18 AM
JonathanEyoon
Integration of Rational Functions by Partial Fractions
Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.

Maybe you guys can help me understand through examples.

2 / (x^2 + 3x - 4)

now I know I can factor the Q(x) to (x + 4)(x - 1)

So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.

Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

1 / (x^6 - x^3)

now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks
• Jun 19th 2008, 09:24 AM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.

Maybe you guys can help me understand through examples.

2 / (x^2 + 3x +4)

now I know I can factor the Q(x) to (x + 4)(x - 1)

So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.

Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

1 / (x^6 - x^3)

now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks

When you have repeated factors you must include a term for each degree of the repeated factor. So for example in your problem

$\displaystyle \frac{1}{x^3(x-1)(x^2+x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^ 3}+\frac{D}{x-1}+\frac{Ex+F}{x^2+x+1}$

The same is true for repeated quadratic factors for example

$\displaystyle \frac{3}{x^2(x^2+9)^2}=\frac{A}{x}+\frac{B}{x^2}+\ frac{Cx+D}{x^2+9}+\frac{Ex+F}{(x^2+9)^2}$

I hope this helps
• Jun 19th 2008, 09:26 AM
Moo
Hello =)

Quote:

Originally Posted by JonathanEyoon
Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.

Maybe you guys can help me understand through examples.

2 / (x^2 + 3x +4)

now I know I can factor the Q(x) to (x + 4)(x - 1)

So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.

Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

1 / (x^6 - x^3)

now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks

For the first one, I guess it is (x^2+3x-4), right ?

When dealing with powers, for example [P(x)]^n, with P(x) a polynomial of degree 1 (that is to say x-4, 2x+3, etc...), we write this :

$\displaystyle \frac{\dots}{[P(x)]^n}=\frac{A}{P(x)}+\frac{B}{[P(x)]^2}+\dots+\frac{N}{[P(x)]^n}$

When you have an irreducible polynomial, such as $\displaystyle x^2+x+1$, then :

$\displaystyle \frac{\dots}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

Good luck (Wondering)

Edit : niaaah >.< too slcow !
• Jun 19th 2008, 09:42 AM
JonathanEyoon
For repeated factors, I must include a term and it goes the same for quadratics. Let me try one on here and tell me if it's correct.

(2x + 1) / ((x + 1)^3)((x^2 + 4)^2)

(A / (x + 1)) + B / ((x + 1)^2) + C / ((x + 1)^3) + Dx +E / (x^2 + 4) + Fx + G / ((x^2 +4)^2)

what do you think? (Worried)
• Jun 19th 2008, 09:45 AM
Moo
Quote:

Originally Posted by JonathanEyoon
For repeated factors, I must include a term and it goes the same for quadratics. Let me try one on here and tell me if it's correct.

(2x + 1) / ((x + 1)^3)((x^2 + 4)^2)

(A / (x + 1)) + B / ((x + 1)^2) + C / ((x + 1)^3) + Dx +E / (x^2 + 4) + Fx + G / ((x^2 +4)^2)

what do you think? (Worried)

I think it's correct (Sun)

Extra feature !!!! Here is what you should get :

$\displaystyle \frac{-(38*x-8)}{625*(x^2+4)}-\frac{21*x+19}{125*(x^2+4)^2}+\frac{38}{625*(x+1)} +\frac{6}{125*(x+1)^2}-\frac{1}{25*(x+1)^3}$
• Jun 19th 2008, 09:58 AM
JonathanEyoon
I really appreciate the help guys! I have one more question (Doh)

How am I supposed to know when to Equate for quadratics, linear, and constants?

Example. ( This is kinda ugly )

Example : (-x^3 + 2x - x + 1) = A(x^4 + 2x^2 + 1) + B(x^4 + x^2) + C(x^3 + x) + Dx^2 + Ex.

• Jun 19th 2008, 10:11 AM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
I really appreciate the help guys! I have one more question (Doh)

How am I supposed to know when to Equate for quadratics, linear, and constants?

Example. ( This is kinda ugly )

Example : (-x^3 + 2x - x + 1) = A(x^4 + 2x^2 + 1) + B(x^4 + x^2) + C(x^3 + x) + Dx^2 + Ex.

I think the 2nd term should be $\displaystyle 2x^2$

First we multiply out the right hand side to get

$\displaystyle -x^3+2x^2-x+1=Ax^4+2Ax^2+A+Bx^4+Bx^2+Cx^3+Cx+Dx^2+Ex$

Now we group all terms of the same degree (same exponent) to get

$\displaystyle -x^3+2x^2-x+1=Ax^4+Bx^4+Cx^3+2Ax^2+Bx^2+Dx^2+Cx+Ex+A$

Now we can factor the right hand side to get

$\displaystyle -x^3+2x^2-x+1=(A+B)x^4+(C)x^3+(2A+B+D)x^2+(C+E)x+A$

Now we can equate the coeffients from both sides of the equation. For exapmle since there is no $\displaystyle x^4$ term on the left hand side its coeffeint must be zero so... $\displaystyle 0=A+B$ now for the cubic term$\displaystyle -1=C$ and so on $\displaystyle 2=2A+B+D$ and $\displaystyle -1=C+E$ finally $\displaystyle 1=A$

Now we can solve this system to get A=1, B=-1,C=-1,D=1,E=0
• Jun 19th 2008, 10:33 AM
JonathanEyoon
ah Ic. Thanks alot! I'm doing the homework for these types of problems at the moment. Will let you guys know if I have any questions.