Given the following Integral
$\int \frac{2x^3+ 2x^2+ 2x+ 1}{x^2 (x^2+1)}$
I would expand my fractions like the following
$\frac{A}{x} + \frac{B}{x^2}+\frac{Cx+D}{x^2+1}$
When I look at the steps for distributing the coefficients
It shows that A is still distributed to x and B isn't distributed to x or $x^2$
This is how they distributed the coefficients on emathhelp and symbolab
A(x)($x^2$+1)+ B($x^2$+1) + Cx+D($x^2$)
The way I was going to write it was
A($x^2$)($x^2$+1)+ B(x)($x^2$+1) + Cx+D(x)($x^2$)
Why is the former the correct way? What I usually do is match each coefficient with the denominator term that it's missing (like the latter). Why does the A distribute to x but B doesn't distribute to $x^2$?
Edit: Is it because they Solved the expression by multiplying the LCD on both sides of the equation?
Also their solution for the new A values are Ax + A$x^3$. That's the correct way to distribute A(x)($x^2$+1)?
After you distribute to A to x you take Ax and distribute it to $x^2$+1 ?
So you said that the form is $\frac{A}{x} + \frac{B}{x^2}+\frac{Cx+D}{x^2+1}$. This isn't.You must note that for every irreducible, quadratic factor in the denominator, there will be a partial fraction in the form of $\frac{Ax+B}{ax^2+bx+c}$. In your case there are 2 of these factors: $x^2$ and $x^2+1$. Just note that there should be any linear denominators.
This should hopefully help.
Multiply both sides by the denominator of the LHS, $x^2(x^2+1)$. Make sure you simplify when necessary and you should get the correct form.