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

So this is the integral and I have been stuck on it for ages without getting anywhere at all. Nothing I tried has gotten me anywhere so I'm basically stuck on nowhere with this.

I would really love for someone to go through this step by step, that's how I learn the best way. If it is even solvable by a human being....

user135839
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  • @Mathmo123. I hope that with $4x^5+1$, the OP would not have used words such as "nearly impossible integral" or "even solvable by a human being" ! Cheers :) – Claude Leibovici Jun 24 '14 at 13:07
  • Well I am fairly new to calculus and I am not as experienced as you might be – user135839 Jun 24 '14 at 13:08
  • What I was commenting is that I am sure that, if the numerator had been $4x^5+1$, I am sure you would have recognized $\frac {u'}{u^2}$. Don't worry about being new ! We all have been and we are here to help you. – Claude Leibovici Jun 24 '14 at 13:13
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    $$\frac{4x^5 - 1}{(x^5+x+1)^2} = \frac{(5x^4+1)x-(x^5+x+1)}{(x^5+x+1)^2} =\left(\frac{-1}{x^5+x+1}\right)'x + \frac{-1}{x^5+x+1}x' = \frac{d}{dx}\left(\frac{-x}{x^5+x+1}\right) $$ – achille hui Jun 24 '14 at 13:30
  • @achillehui: Isn't that like working backwards? :P Those steps weren't obvious to me until I checked the final answer from W|A. – Pranav Arora Jun 24 '14 at 13:33
  • @PranavArora it is not obvious but it isn't something you must need a CAS to figure out. – achille hui Jun 24 '14 at 13:35
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    @achillehui: Yes, you are right indeed, these type of problems shouldn't require a CAS but converting an integrand to the form $\dfrac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}$ is never obvious to me. I always need to seek alternative methods. :( – Pranav Arora Jun 24 '14 at 13:38

3 Answers3

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Hint

Notice that the denominator is a square. So suppose that $$\frac{a+b x}{x^5+x+1}$$ is the antiderivative of $$\frac{4x^5 -1} {(x^5 + x +1)^2}$$ Differentiate the first expression and identify the terms.

I let you finding why the numerator cannot be more complex.

I am sure that you can take from here.

Claude Leibovici
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    I don't really understand what you mean. – user135839 Jun 24 '14 at 13:05
  • Should i use partial fraction as method? – user135839 Jun 24 '14 at 13:09
  • OK. Admit that I may have given the trick. So, compute the derivative of my expression and search for which values of $a$ and $b$ you obtain your integrand. – Claude Leibovici Jun 24 '14 at 13:10
  • No partial fractions. Try what I suggest in my comment and come back. – Claude Leibovici Jun 24 '14 at 13:11
  • Did you actually try it yourself? It doesn't seem to work out nicely, and the answer given by Maple doesn't agree either. It looks rather messy, in fact. – Harald Hanche-Olsen Jun 24 '14 at 13:26
  • @HaraldHanche-Olsen. For sure, I did it (by hand) and the answer is very simple. No need of any CAS. – Claude Leibovici Jun 24 '14 at 13:30
  • Argh, you're right: In my calculation, $a+bx$ morphed into $ax+b$ while its derivative remained as $b$. It seems that $ax+b$ is the canonical first order polynomial in my head. It is interesting though, that Maple came up with a complicated answer containing two sums of logarithms, summed over the roots of two different third degree polynomials. These sums must cancel somehow. I guess it shows that CASes can be deceptive. – Harald Hanche-Olsen Jun 24 '14 at 14:00
  • Why can't the numerator be more complex? – A Googler Jul 22 '15 at 09:45
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$$\int \frac{4x^5-1}{(x^5+x+1)^2}\,dx=\int \frac{4x^5+5x^4-5x^4-1}{(x^5+x+1)^2}\,dx$$ $$=\int \frac{4x^5+5x^4}{(x^5+x+1)^2}\,dx-\int \frac{5x^4+1}{(x^5+x+1)^2}\,dx=J_1-J_2$$ $J_2$ can be handled by using the substitution $x^5+x+1=t$ and comes out to be $\dfrac{-1}{x^5+x+1}$. For $J_1$, factor out $x^{5}$ from the expression in parentheses in denominator, i.e: $$J_1=\int \frac{4x^5+5x^4}{(x^5+x+1)^2}\,dx= \int \frac{\dfrac{4}{x^5}+\dfrac{5}{x^6}}{\left(1+\dfrac{1}{x^4}+\dfrac{1}{x^5}\right)^2}\,dx$$ Now use the substitution $1+\dfrac{1}{x^4}+\dfrac{1}{x^5}=t\Rightarrow -\left(\dfrac{4}{x^5}+\dfrac{5}{x^6}\right)\,dx=dt$, hence $$J_1=-\int \frac{dt}{t^2}=\frac{1}{t}+C=\frac{x^5}{x^5+x+1}+C$$ Hence, $$J_1-J_2=\frac{x^5}{x^5+x+1}+\frac{1}{x^5+x+1}+C=\boxed{\dfrac{x^5+1}{x^5+x+1}+C}$$

Pranav Arora
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  • This is very complex, don't you think ? Cheers. – Claude Leibovici Jun 24 '14 at 13:43
  • @ClaudeLeibovici: Nope, I don't think so. Rather, its one of the most standard trick which was taught to me in high school. :) – Pranav Arora Jun 24 '14 at 13:44
  • @ClaudeLeibovici This is complex but correct. Start from the original integrand, there is no apriori reason it has a simple solution. – achille hui Jun 24 '14 at 13:46
  • @PranavArora. As achille hui showed, life is simpler. – Claude Leibovici Jun 24 '14 at 13:46
  • I prefer this method ectually, but I have a hard time understanding the factorization of x^5 of J1. I can see that in the denominator you have devided by x^5 but what happens in the numinator is beyond what I understand. – user135839 Jun 26 '14 at 19:30
  • @user135839: That is simple algebra, I do not see why it is difficult to understand. If you factor out $x^5$ from the expression in parentheses, you get $x^{10}$. Divide the numerator by this $x^{10}$ i.e $4x^5/x^{10}=4/x^5$ and $5x^4/x^{10}=5/x^6$. – Pranav Arora Jun 27 '14 at 04:22
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Ok So I know that this will probably be super wrong but it's what I could come up with

The answere i found was:

$$ -\frac{1 } {x^4}+4log(x+1) -\frac{16x+15}{12x^4}+4x + log(x+1) +4 $$

I used barak manos hint for this

user135839
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