HELP on Partial Fraction Decomposition Problem - Heavy Side Technique

[khatche4]

HELP! on Partial Fraction Decomposition Problem! - Heavy Side Technique

I am doing a partial fraction decomposition problem for my calc 2 class.
We use the Heavy Side Technique, but I will take help either way!

$$\frac{2x-1}{x(x^2+1)^2}$$

Thank you!

 

[gabbagabbahey]

Hi khatche4, welcome to PF!

What have you tried?...You need to show some attempt at the problem before we will assist you.

 

[khatche4]

I tried splitting it up as..

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

Then I used x=0, x=1, x=-1, x=2, and x=-2 to get...

1= -4 + 2B + 2C + 2D + 2E
3= 4 - 2B + 2C - 2D + 2E
3= -25 + 20B + 10C + 20D + 10E
5= 25 - 20B + 10C - 20D + 10E

And then I got stuck...


Should it be this, initially?:

$$\frac{2x-1}{x(x^2+1)^2}$$ = $$\frac{A}{x}$$ + $$\frac{Bx+C}{x^2+1}$$ + $$\frac{Dx+E}{(x^2+1)^2}$$ ?

 

[gabbagabbahey]

Should it be this, initially?:

$$\frac{2x-1}{x(x^2+1)^2}$$ = $$\frac{A}{x}$$ + $$\frac{Bx+C}{x^2+1}$$ + $$\frac{Dx+E}{(x^2+1)^2}$$ ?

Yes, it should.

 

[khatche4]

Ok! thank you! If I need anymore help, I'll come back!
=]

 

[khatche4]

Ok... I'm stuck again.

I went along with
$$\frac{2x-1}{x(x^2+1)^2}$$ = $$\frac{A}{x}$$ + $$\frac{Bx+C}{x^2+1}$$ + $$\frac{Dx+E}{(x^2+1)^2}$$

From there, I used
x=0 to get A = -1
x=1
x=-1
x=2
x=-2
(Am I supposed to use 4?? We've only done up to Bx+C, and there we use 2.. So I assume to use 4)..

Then, from there, I got
1 = -4 + 2B + 2C + D + E
3 = 4 - 2B + 2C - D + E
3 = -25 + 20B + 10C + 4D + 2E
5 = 25 - 20B + 10C - 4D + 2E

So that boils down to..

1 = 2C + E
3 = 2C + E
3 = 10C + 2E
5 = 10C + 2E
right??

Will I be using substitution for this one? If so, do I use 2 different substitutions or just one?

Thanks, again!

 

[gabbagabbahey]

Ok... I'm stuck again.

I went along with
$$\frac{2x-1}{x(x^2+1)^2}$$ = $$\frac{A}{x}$$ + $$\frac{Bx+C}{x^2+1}$$ + $$\frac{Dx+E}{(x^2+1)^2}$$

From there, I used
x=0 to get A = -1

good

x=1
x=-1
x=2
x=-2
(Am I supposed to use 4?? We've only done up to Bx+C, and there we use 2.. So I assume to use 4)..

You can get away with using only 3 values, if you choose those values wisely. What made x=0 such a useful choice was that it made one of the factors of $x(x^2+1)^2$ zero...The other factor is of course just $x^2+1$, which is zero at $x=\pm i$, so try using $x=i$ to find easy solutions for $D$ and $E$; you'll have a much easier time finding $B$ and $C$ after that.