# Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$)

I need to simplify $$\begin{equation} \frac{x^{5}}{x^{2}+1} \end{equation}$$ by long division in order to solve an integral. However, I keep getting an infinite series: $$\begin{equation} x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+... \end{equation}$$

• In polynomial long division, we need to divide until the degree of remainder became less than degree of divisor, if we continue it further then quotient will no longer be a polynomial. – user629353 Jan 9 '19 at 8:39

## 6 Answers

When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $$x^3+x+1$$ by $$x+1$$,

• First multiply $$x+1$$ by $$x^2$$ and subtract the result, $$x^3+x^2$$, from $$x^3+x+1$$, giving the remainder $$-x^2+x+1$$;

• Now, multiply $$x+1$$ by $$-x$$ and subtract the result, $$-x^2-x$$, from $$-x^2+x+1$$, giving the remainder $$2x+1$$;

• As a final step, multiply $$x+1$$ by $$2$$ and subtract the result, $$2x+2$$, from $$2x+1$$, giving you the remainder $$-1$$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$

Note that you don't go beyond the last step, because the degree of the remainder, $$2,(0)$$ is smaller than the degree of the divisor, $$x+1, (1)$$. To go any further, you will have to multiply $$x+1$$ by terms containing negative powers of $$x$$, which would mean the quotient will no longer be a polynomial in $$x$$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.

The essence of performing long-division to solve the integral $$\int\frac{x^5}{x^2+1}dx$$ is to express $$x^5=(x^2+1)Q(x)+R(x)$$, where $$Q(x),R(x)$$ are polynomials in $$x$$ with degree of $$R(x)<2$$, which is the degree of the divisor, $$x^2+1$$.

This gives $$\frac{x^5}{x^2+1}=Q(x)+\frac{R(x)}{x^2+1}$$where $$Q(x)$$ can be easily integrated because it is a polynomial, and $$\displaystyle\frac{R(x)}{x^2+1}$$ can be integrated using partial fractions or similar techniques.

As other answers have pointed out, $$\int\frac{x^5}{x^2+1}dx=\int x^3-x+\frac x{x^2+1}dx=\frac{x^4}4-\frac{x^2}2+\frac12\ln(x^2+1)+C$$

The idea of polynomial division is like integer division. With integer division of $$\frac nd$$, we want integer $$q,r$$ so that $$n=qd+r$$ and $$r\lt d$$. With polynomial division of $$\frac nd$$, we want polynomial $$q,r$$ so that $$n=qd+r$$ and $$\deg(r)\lt \deg(d)$$. $$\require{enclose} \begin{array}{rl} &\phantom{)\,}\color{#C00}{x^3}\color{#090}{-x}\\[-4pt] x^2+1\!\!\!\!\!&\enclose{longdiv}{x^5\qquad}\\[-4pt] &\phantom{)\,}\underline{\color{#C00}{x^5+x^3}}\\[-2pt] &\phantom{)\,x^5}{}-x^3\\[-4pt] &\phantom{)\,x^5}\underline{\color{#090}{{}-x^3-x}}\\[-4pt] &\phantom{)\,x^5{}-x^3-{}}x\\[-4pt] \end{array}$$ So we get a quotient of $$x^3-x$$ and a remainder of $$x$$, which allows us to write both $$\overbrace{\quad\,x^5\quad\,}^n=\overbrace{\left(x^3-x\right)}^q\overbrace{\left(x^2+1\right)}^d+\overbrace{\vphantom{x^5}\quad\;x\quad\;}^r$$ and $$\frac{x^5}{x^2+1}=x^3-x+\frac{x}{x^2+1}$$

Why wouldn't you get an infinite series?

If $$x^2 + 1$$ doesn't divide evenly into $$x^5$$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.

The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.

$$\frac {x^5}{x^2+ 1} = \frac {x^5 + x^3}{x^2 + 1} -\frac {x^3}{x^2 + 1}$$

$$= x^3 - \frac {x^3 + x}{x^2 + 1} + \frac {x}{x^2+ 1} =$$

$$x^3 - x + \frac {x}{x^2 + 1}$$.

Now we can't divide any further as the degree of the denominator ($$x^2 + 1$$) is $$2$$ and that is larger than the degree of that numerator ($$x$$). So we are done.

$$\frac {x^5}{x^2 +1} = x^3 - x +\frac {x}{x^2 +1}$$.

Put another way: $$x^5 = (x^3 - x)(x^2 + 1) + x$$.

$$x$$ is .... just a remainder you cant do any thing with.

It is exactly like.

$$\frac {249}{7} = \frac {210 + 39}{7} = \frac {210}7 + \frac {39}7=$$

$$30 + \frac {35 + 4}{7} = 30 + \frac {35}7 + \frac 47=$$

$$30 + 5 + \frac 47 = 35\frac 47$$.

We've divided as far as we can go.

If you tried to go further we would get decimals:

$$30 + 5 + \frac {40}{7*10} = 30 + 5 + {35 + 5}{70} =$$

$$30 + 5 + \frac 5{10} + \frac 5{70} =30 + 5 + \frac 5{10} + \frac {50}{700} =$$

$$30 + 5 + \frac 5{10} + \frac 7{100} + \frac 1{1000} + ......$$.

$$= 35.571428571428571428571428571429.....$$

But we weren't asked to go that for and as we aren't masochists.... we stopped at $$35\frac 47$$.

\$

$$\frac {x^{5}} {x^{2}+1}= x^{3}-x+\frac x {x^{2}+1}$$.

Since $$x^5 = \left(x^2 + 1\right)\left(x^3 - x\right) + x$$, we have that

$$\cfrac{x^5}{x^2 + 1} = x^3 - x + \cfrac{x}{x^2 + 1} \tag{1}\label{eq1}$$

To do this in general, first note that $$x^2$$ divides into $$x^5$$ a total of $$x^3$$ times. However, this gives $$x^3\left(x^2 + 1\right) = x^5 + x^3$$, so it's too big by $$x^3$$. As such, you need to subtract an appropriate value, with it being $$x$$ here due to $$x \times x^2 = x^3$$. However, $$x\left(x^2 + 1\right) = x^3 + x$$, so subtracting this means that you are now $$x$$ too small, so you need to add that $$x$$ back. However, as the degree of $$x$$ is only $$1$$, which is less than the degree of $$2$$ in $$x^2 + 1$$, you leave that final $$x$$ as the remainder. This, in total, gives \eqref{eq1}.

If the integral is well defined, you can write $${x^5\over 1+x^2}dx={1\over 2}{(x^2)^2\over 1+x^2}dx^2={1\over 2}{u^2\over 1+u}du$$