What is the integral of $\frac{x^7}{x^3+1}$?

Question

How to find this integral? $$\int \frac{x^7}{x^3+1}\,dx$$

I think it may need a partial fraction but I'm not sure. Just need a start. Any help would be appreciated.

Answer 1

Yes! Partial fractions is definitely the way to go.

There is a lot of ways to think about partial fractions, so hopefully what I say is generic enough to sound familiar.

The first thing is that you need to make sure the numerator doesn't have bigger (or equal) degree. Since $7 \geq 3$, you need to do polynomial division. This is where you divide $x^3 + 1$ into $x^7$. The result will be a quotient $g(x)$ (which should have degree $4$) and a remainder $f(x)$, see for example

http://library.wolfram.com/webMathematica/Education/LongDivide.jsp

Next you will have a new fraction of the form $\frac{f(x)}{x^3+1}$ where the degree of $f$ is strictly less than $3$. When I did this I got $f(x) = x$. Make sure you know how to do this part! It is very useful algebra and comes up a lot in partial fraction decompositions. In summary what we have just done is

$\frac{x^7}{x^3+1} = g(x) + \frac{f(x)}{x^3+1} = g(x) + \frac{x}{x^3+1}$

Now you probably know how to integrate $g(x)$ so we will skip that bit.

For the last one, you need to split up the denominator as much as you can. You can quickly guess/check that $-1$ is one solution to $x^3+1$. That means $x+1$ is a factor of $x^3+1$. If you repeat the polynomial division again you'll find

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

so in particular

$x^3+1 = (x+1)(x^2-x+1)$

Now things should start looking familiar. The current goal right now is to solve

$\int\frac{x}{(x+1)(x^2-x+1)}dx$

And here you can use the usual partial fraction process. That is, because we have a linear and a quadratic term on the denominator we introduce some variables $A,B,C$ and write

$\frac{x}{(x+1)(x^2-x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}$

then use whatever method you want to find $A,B,C$ (equating coefficients/limit arguments/etc).

The linear term with the $A$ in it is an easy integral. The quadratic one you can use a trick such as completing the square in order to either use a trig sub or a $u$-sub. I will leave that to you.

I think the most important thing is that all these steps have been algebra! We didn't really do any calculus (that's the part I left out). This shows how important algebra skills are for calculus.

Answer 2

By polynomial long division, we get $$\int \frac{x^7}{x^3 + 1}\,dx = \int \left(x^4 - x + \frac{x}{x^3 + 1}\right)\,dx = \int \left(x^4 - x + \frac{x}{(x+1)(x^2 - x+1)}\right)\,dx$$

The last term in the integral can be integrated using partial fraction decomposition: $$\frac{x}{(x+1)(x^2 - x+1)} = \frac A{x+1} + \frac{Bx+C}{x^2-x+1}$$

where $A(x^2 - x+1) + (Bx+C)(x+1) = x\tag{*}$

Let's first put $x=-1$ to determine $A$: $3A = -1 \iff \color{blue}{A = -\dfrac 13}$

Now we expand $(*)$: $$(A+B) x^2 + (B+C-A )x + (A+C)= x$$ So $$(A+B) = 0 \implies \color{blue}{B = \frac 13}, \quad (B+C-A) = 1 \implies \color{blue}{C=\frac 13},$$ and as a check, $C+A = 0 \iff \frac 13 - \frac 13 = 0 \checkmark$