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.