# Method of Partial Fractions integration

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

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

???

• 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 $\frac13\big[\frac1y-\frac1{y+3}\big]=\frac13\big[\frac1{x^2}-\frac1{x^2+3}\big]$ Mar 12 '19 at 21:13
• Wow, we've got the equal-isn't-equal-to-equal-unless-when-it-is-equal-to-equal kind of equal in the equation. Mar 12 '19 at 21:19

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.

• Oh, yes you are totally right. However that didn't seem to change my system of equations outcome still. Please see my edits above Mar 12 '19 at 21:35
• 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 $A=C=0$, $B=1/3$, $D=-1/3$.
– A.M.
Mar 12 '19 at 21:56
• How are you isolating A,B, and C to solve for them? Is there a more details you can provide or some additional resources? Mar 12 '19 at 22:02
• 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) Mar 12 '19 at 23:26
• 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.
– A.M.
Mar 12 '19 at 23:46

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 $$D = -B = -\frac{1}{3}$$.