How to express $\frac{x^3+4x^2-1}{(x^2+1)^2}$ as a polynomial plus a proper fraction, using long division? I'm trying to express
$$\dfrac{x^3+4x^2-1}{(x^2+1)^2}$$
as a polynomial plus a proper fraction, using long division but I don't know how to do that. It'd be cool if you can solve this. Thanks.
 A: If $f(x)$ and $g(x)$ are two polynomials, $g(x)\neq 0$, then there exist unique polynomial $q(x)$ and $r(x)$, called the "quotient" and the "remainder", such that
$$\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)},\qquad r(x)=0\text{ or }\deg(r)\lt\deg(g).$$
Both $q$ and $r$ can be found using polynomial long division. 
(We often refer to a fraction of polynomials in which the degree of the numerator is strictly smaller than the degree of the denominator as a "proper fraction", in analogy to the case of numerical fractions $\frac{a}{b}$, which are "proper" when $|a|\lt |b|$, and improper if $|a|\geq|b|$; every fraction $\frac{a}{b}$ with $a$ and $b$ integers, $b\neq 0$, can be written uniquely as $\frac{a}{b} = n + \frac{r}{b}$, where $n$ and $r$ are integers and $0\leq r\lt|b|$; this is the analogous operation with polynomials). 
In your case, since $f(x) = x^3 + 4x -1$ and $g(x) = (x^2+1)^2$, we already have $\deg(f)\lt \deg(g)$, so the fraction is already proper; that means that $q(x) = 0$ and $r(x) = f(x)$, and there is nothing left to do.
A: Hint $\rm\ \ x^3 + 4\: x^2 - 1\ =\ (x+4)\:(x^2+1) - (x+5)\ $ by the Polynomial division algorithm.
Now divide both sides by $\rm\:(x^2+1)^2\:$ and cancel a factor of $\rm\:x^2+1$ from one term.
A: By the rational root theorem, you know the only possible rational roots would be $\pm1$, neither of which is in fact a root.
The numerator factors (approximately) as
$$
(x+3.93543233197003)(x+0.537401577025226)(x-0.472833908995256)
$$
which are irrational algebraic roots.
You could use the cubic equation to find the closed form expression for these.
