Integral and Partial fractions

[jimen113]

Homework Statement

use the method of partial fractions on \int \frac{36}{(x-2)(x-1)^2(x+1)^2} dx

2. Homework Equations




3. The Attempt at a Solution

\frac{A}{(x-2)} + \frac{B}{(x-1)} + \frac{C}{(x-1)^2} + \frac{D}{(x+1)} + \frac{E}{(x+1)^2}
I took the 36 out of the integral. Then

A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1
In the first fraction (A) (all terms cancel)
I tried to plug in x=
1, -1, 0, 2, -2, into (A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1)
so...
when x=1 I ended up with B(-2) +C(-4)=1 (I know this is probably simple algebra or I'm definitely screwing up somewhere

Any help is appreciated

 

[Cyosis]

Why do you have this extra factor (x+1) in front of everything? For example for A you would get:

\frac{A}{x-2} (x-2)(x-1)^2(x+1)^2=A (x-1)^2(x+1)^2. You have added an extra (x+1) and you're doing that for every term it seems.

You start out with:

\frac{A}{x-2}+\frac{B}{x-1}+\frac{C}{(x-1)^2}+\frac{D}{x+1}+\frac{E}{(x+1)^2}=\frac{1}{(x-2)(x-1)^2(x+1)^2}

right?

Then multiply both sides with the denominator of the right hand side.

A(x-1)^2(x+1)^2+B(x-2)(x-1)(x+1)^2+C(x-2)(x+1)^2+D(x-2)(x-1)^2(x+1)+E(x-2)(x-1)^2=1

Now set x=1 \Rightarrow C=-\frac{1}{4}, x=2 \Rightarrow A=\frac{1}{9} etc.

 

[zcd]

You made the mistake over here:

A(x-1) (x-1)^2 (x+1) (x+1)^2 + B(x-2) (x-1)^2 (x+1) (x+1)^2 + C (x-2)(x-1) (x+1) (x+1)^2 + D(x-2) (x-1) (x-1)^2 (x+1)^2 + E (x-2)(x-1) (x-1)^2 (x+1) =1

Because you multiple the separate fractions by the common denominator, it would end up as

A(x-1)^2 (x+1)^2 + B(x-2)(x-1)(x+1)^2 + ... + E(x-2)(x-1)^2 = 1

 

[jimen113]

Thank you so much for your help zcd and cyosis. Last question: I get A= (1/9) C=(-1/4)
E=(-1/12)
But to find D and B I tried plugging in zero and didn't get anywhere. What other numbers can I plug into solve for D and E?

 

[Cyosis]

You will have to pick two values such that B and D aren't multiplied by zero. Since you know all the other constants you will get two equations relating B to D. Solve them and you're good to go.

 

[jimen113]

omg...thank you for your help...
I did try plugging in (-2) and I end up with:
1+B(12)+1+D(36)+3=1
and B(12)+D(36)=-4
so I'm still left with the two unknowns...

 

[Cyosis]

Try x=0 this will give you 1/9+2B+1/2-2D+1/6=1, -> B-D=1/9. You now have two equations and two unknowns so it is solvable.

 

[jimen113]

thanks.