Integrating (using partial fractions) Apostol Section 6.25 #25

[process91]

Homework Statement

$$\int\frac{4x^5-1}{(x^5+x+1)^2}dx$$

Homework Equations

This is in the section on Partial Fractions. The main idea in this section was that you get the integral down to a sum integrals of the following forms:
$$\int\frac{dx}{(x+a)^n} , \int \frac{x dx}{(x^2 + bx + c)^m} , \int \frac{dx}{(x^2+bx+c)^m}$$

The Attempt at a Solution

The basic approach for most of these was to just use partial fractions by factoring the denominator and algebraically breaking down the result. I factored the denominator to

$$\left(x^2+x+1\right)^2\left(x^3-x^2+1\right)^2$$

The term on the right could be factored again, but it doesn't look promising.


I sense I should be using a different approach with this problem, but I'm not sure what.

Please just give me a hint, if possible.

 

[micromass]

Uuh, here's a way to solve it without using partial fractions. Try to write
$$\frac{4x^5-1}{(x^5+x+1)^2}$$

in the form of

$$\frac{f^\prime g-g^\prime f}{g^2}$$

Try to determine f here.


But how to do it with partial fractions?? Well, the only way I see is to write

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

and proceed formally. The hope is that the $\alpha$ will cancel itself in the end (which it will here). But this is likely to become very difficult.

 

[process91]

The quotient rule idea is what I came to after seeing the answer, and after I saw it I couldn't un-see it. Many of the other problems in this section have been extremely cumbersome arithmetically for me, so it's very possible that the $\alpha$ method would be what Apostol had in mind.

 

[SammyS]

Uuh, here's a way to solve it without using partial fractions. Try to write
$$\frac{4x^5-1}{(x^5+x+1)^2}$$

in the form of

$$\frac{f^\prime g-g^\prime f}{g^2}$$

Try to determine f here.
...

It makes sense to assume that f(x) is a polynomial & of course g(x) = x⁵ + x + 1. After some consideration you can convince yourself that f(x) is of the form, f(x) = ax + b.

WolframAlpha gets the surprisingly uncomplicated answer, but I couldn't get it to show the steps.