0
$\begingroup$

$\int \frac{dx}{x^4+3x^2} = \int \frac{dx}{x^2(x^2+3)} = A(x^2+3) + (Bx+C)x^2 = \frac{A}{x^2} + \frac{Bx+C}{x^2+3} = Ax^2 + 3A + Bx^2 + Cx^2$

I am having some trouble solving the system of equations that follows. I tried plugging into my calculator a 4x4 matrix $\begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 3 & 0 & 0 & 0 \\ \end{bmatrix}$ but the RREF form only gave me $A,B,C = 0$ which doesn't seem like the right answer and I am now stuck.

I would prefer hints rather than answers at this time!


Edit:

My initial partial fractions was incomplete. The correct one is posted below:

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

$$ = Ax^3 + 3Ax + Bx^2 + 3B + Cx^3 + Dx^2 $$

This gives me the following matrix:

$ \begin{bmatrix} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 3 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 \\ \end{bmatrix}$ but the RREF is still giving me $A,B,C,D = 0$!


Edit: attempting to solve system of equations...

$1 = Ax(x^2+3) + B(x^2+3) + Cx^3 + Dx^2$ Let x = 1,

1 = A(4) + B(4) + C + D

???

$\endgroup$
2
  • 1
    $\begingroup$ You have a product of two quadratics in the denominator with no linear term. Make the substitution $x^2=y$ (only to find the partial fraction decomposition and not for integration). The integrand becomes $\frac1{y(y+3)}$ which is a product of linear terms and can be easily decomposed into $\frac13\big[\frac1y-\frac1{y+3}\big]=\frac13\big[\frac1{x^2}-\frac1{x^2+3}\big]$ $\endgroup$ Mar 12 '19 at 21:13
  • 1
    $\begingroup$ Wow, we've got the equal-isn't-equal-to-equal-unless-when-it-is-equal-to-equal kind of equal in the equation. $\endgroup$ Mar 12 '19 at 21:19
5
$\begingroup$

For partial fractions involving denominators of perfect powers, you need to split it up in a slightly different way:

$$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+3}$$

Edit: in this specific case, we can get the equality $$1 = Ax(x^2+3) + B(x^2+3) + Cx^3 + Dx^2$$

From this we can obtain a set of simultaneous equations by substituting $x = 0,1,-1,2$ to get:

$$ 1 = 3B $$ $$ 1 = 4A+4B+C+D $$ $$ 1 = -4A+4B-C+D $$ $$ 1 = 14A+7B+8C+4D $$

You can solve this with any method you like (e.g. RREF).

Double edit: I've just worked out the way you were taught to do it, by comparing coefficients rather than making substitutions. This works too. The issue is simply in your matrix: you have that you want $Ax^3 + 3Ax + Bx^2 + 3B + Cx^3 + Dx^2$ to be equal to $1$ (the numerator of the original fraction). This is why it's important to take care of where you use equality signs: your string of equalities in the original post are not correct. Your matrix should be

$$ \begin{bmatrix} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 3 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 \\ \end{bmatrix} $$

Note the $1$ in the bottom right corresponding to the $1$ in the numerator of the initial fraction. Solving this will give you $A=C=0$, $B=1/3$, $D=-1/3$, which is correct. And then you can integrate.

$\endgroup$
6
  • $\begingroup$ Oh, yes you are totally right. However that didn't seem to change my system of equations outcome still. Please see my edits above $\endgroup$
    – Evan Kim
    Mar 12 '19 at 21:35
  • $\begingroup$ I'm not familiar with the matrix way of doing this. Here's how I learned it: we know $\frac{1}{x^2(x^2+3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+3}$ so multiply upwards by $x^2(x^2+3)$ and you get that $1 = Ax(x^2+3) + B(x^2+3) + Cx^3 + Dx^2$. Then by putting $x=0,1,-1,2$ or any other convenient values, you get simultaneous equations which should yield $A=C=0$, $B=1/3$, $D=-1/3$. $\endgroup$
    – A.M.
    Mar 12 '19 at 21:56
  • $\begingroup$ How are you isolating A,B, and C to solve for them? Is there a more details you can provide or some additional resources? $\endgroup$
    – Evan Kim
    Mar 12 '19 at 22:02
  • $\begingroup$ please see my new edit, I don't think I understand how to solve the system of equations after letting $x=1$ and how to "get the simultaneous equations" (not sure why you call it simultaneous equations) $\endgroup$
    – Evan Kim
    Mar 12 '19 at 23:26
  • 1
    $\begingroup$ See my edit. Do what you did but not just for $x=1$; also for $x=0$, $x=-1$ and $x=2$. The values of $x$ chosen are arbitrary - the point is that with each substitution you get a new equation, and for four unknowns you need four equations. $\endgroup$
    – A.M.
    Mar 12 '19 at 23:46
1
$\begingroup$

Even in the revised decomposition equation in the question, there is an error: The denominator is missing on the right-hand side: We should have $$\frac{1}{x^2 (x^2 + 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx+D}{x^2+3} .$$

Hint Before proceeding with cross-multiplication or substitution, we can simplify the resulting linear algebra by observing that the rational function that we are decomposing (on the left-hand side) is even, hence the right-hand side must be too. This immediately forces $$A = C = 0 .$$ (To see this, recall that since an even function is unchanged by the replacement $x \mapsto -x$, applying that substitution to the right-hand side and comparing with the original equation gives those values.)

With this in hand, the decomposition equation simplifies to $$\frac{1}{x^2 (x^2 + 3)} = \frac{B}{x^2} + \frac{D}{x^2+3}. $$

Cross-multiplying gives $$1 = B(x^2 + 3) + D(x^2) = (B + D)x^2 + 3 B,$$ comparing the constant coefficients gives $3B = 1$, so $B = \frac{1}{3}$, and then comparing the remaining coefficients gives $B + D = 0$, so $D = -B = -\frac{1}{3}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.