Partial fraction decomposition for .

[itachi8]

Partial fraction decomposition for...

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

Homework Equations

3. A/x + Bx + C/ x^2 +4

after multiplying through by the denominator and my attempt at finding A,B,C i get this:

3x-1/x(x^2+4) = - one fourth / x + one fourth + 3/ x^2 +4. I don't feel comfortable about this answer.

 

[median27]

The answer i get is:
-(1/4)ln(x)-(1/8)ln(x^2+4)+(3/2)arctan(x/4)+c

i think you made a mistake at Bx which is supp0sed to be B(2x)/(x^2+4).

 

[eumyang]

@itachi: Please either learn LaTeX or use parentheses. I originally read this:

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

as this:
$$3x - \frac{1}{x(x^2 + 4)}$$

Then I originally read this:

3. A/x + Bx + C/ x^2 +4

as this:
$$\frac{A}{x} + Bx + \frac{C}{x^2 + 4}$$

And finally your answer:

3x-1/x(x^2+4) = - one fourth / x + one fourth + 3/ x^2 +4. I don't feel comfortable about this answer.

I thought was this:
$$-\frac{1/4}{x} + \frac{1}{4} + \frac{3}{x^2} + 4$$

But it looks like you meant this as your answer:
$$-\frac{1/4}{x} + \frac{\frac{1}{4} + 3}{x^2 + 4}$$

If so, you forgot an x next to the 1/4 in the 2nd fraction.

@median27, I'm not sure how you got your answer.

 

[median27]

Disregard my post. I apply integration. I thought your query about partial fractions was under integral calculus. I've never encounter partial fraction decomposition in other subjects other than in integral so i assumed it as a query about integration. :D

 

[SammyS]

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

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

Homework Equations

3. A/x + Bx + C/ x^2 +4

A/x + (Bx + C)/(x^2 +4)

Parentheses are important !

Show the rest of your work (with correct grouping) so we can see what you've done.

 

[itachi8]

sorry about the mistake i made with the problem. yes the problem is (3x-1)/(x(x^2 +4)). From the equation i came up with A/x + Bx + C/ x^2 +4. After multiplying through the equation by the denominator (x(x^2+4)) i get A(x^2 +4) + (Bx +C)(x). That gives me Ax^2 + 4A + Bx^2 + Cx. After equating coefficients i come up with A + B=0, C=3, and 4A=-1. I solve for 4A which gives me A= -1/4 and plug this into A + B=0 which gives me B= 1/4. This leads me to my answer which is (- one fourth / x) + (one fourth x + 3/ x^2 +4). Is this correct?

 

[eumyang]

sorry about the mistake i made with the problem. yes the problem is (3x-1)/(x(x^2 +4)). From the equation i came up with A/x + Bx + C/ x^2 +4.

Don't mean to be so picky, but you're still not applying enough parentheses. This looks like
$$\frac{A}{x} + Bx + \frac{C}{x^2} + 4$$.
Without LaTeX, you should have typed
A/x + (Bx + C)/(x^2 +4)

After multiplying through the equation by the denominator (x(x^2+4)) i get A(x^2 +4) + (Bx +C)(x). That gives me Ax^2 + 4A + Bx^2 + Cx. After equating coefficients i come up with A + B=0, C=3, and 4A=-1. I solve for 4A which gives me A= -1/4 and plug this into A + B=0 which gives me B= 1/4. This leads me to my answer which is (- one fourth / x) + (one fourth x + 3/ x^2 +4). Is this correct?

And your answer looks like
$$-\frac{1/4}{x} + \frac{1}{4}x + \frac{3}{x^2} + 4$$

But yes, I am getting the same values for A, B, and C.


EDIT: Wow, 500 posts already?

 

[Ray Vickson]

sorry about the mistake i made with the problem. yes the problem is (3x-1)/(x(x^2 +4)). From the equation i came up with A/x + Bx + C/ x^2 +4. After multiplying through the equation by the denominator (x(x^2+4)) i get A(x^2 +4) + (Bx +C)(x). That gives me Ax^2 + 4A + Bx^2 + Cx. After equating coefficients i come up with A + B=0, C=3, and 4A=-1. I solve for 4A which gives me A= -1/4 and plug this into A + B=0 which gives me B= 1/4. This leads me to my answer which is (- one fourth / x) + (one fourth x + 3/ x^2 +4). Is this correct?

Is A/x + Bx + C/x^2+4 supposed to be A/x + Bx + 4 + C/x^2, or is it A/x + Bx + C/(x^2+4)? You don't need to use LaTeX (although it would be better if you did), but you should use brackets. Why wouldn't you? It is easy and avoids confustion.

RGV