Fdtxjr

$int fracdxx^4+3x^2 = int fracdxx^2(x^2+3) = A(x^2+3) + (Bx+C)x^2 = fracAx^2 + fracBx+Cx^2+3 = Ax^2 + 3A + Bx^2 + Cx^2$

I am having some trouble solving the system of equations that follows. I tried plugging into my calculator a 4x4 matrix
$beginbmatrix
0 & 1 & 0 & 0 \
1 & 0 & 1 & 0 \
0 & 0 & 0 & 0 \
3 & 0 & 0 & 0 \
endbmatrix$ 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:

$$fracAx + fracBx^2 + fracCx+Dx^2+3
= A(x)(x^2+3) + B(x^2+3) + (Cx+D)x^2$$

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

This gives me the following matrix:

$
beginbmatrix
1 & 0 & 1 & 0 & 0 \
0 & 1 & 0 & 1 & 0 \
3 & 0 & 0 & 0 & 0 \
0 & 3 & 0 & 0 & 0 \
endbmatrix$ 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

???

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

$$fracAx + fracBx^2 + fracCx+Dx^2+3$$

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

$$ beginbmatrix
1 & 0 & 1 & 0 & 0 \
0 & 1 & 0 & 1 & 0 \
3 & 0 & 0 & 0 & 0 \
0 & 3 & 0 & 0 & 1 \
endbmatrix $$

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.

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
$$frac1x^2 (x^2 + 3) = fracAx + fracBx^2 + fracCx+Dx^2+3 .$$

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
$$frac1x^2 (x^2 + 3) = fracBx^2 + fracDx^2+3. $$

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 $B = frac13$, and then comparing the remaining coefficients gives $B + D = 0$, so $D = -B = -frac13$.