how to evaluate $ \int \frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2} dx$?

Question

I saw the following problem $$ \int \frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2} dx$$

I tried to solve it by partial fractions and after 7 pages of a lot of calculations I was able to prove that $ \frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2} =\frac{2x+1}{x^2+x+1}+ 2\frac{2x+1}{(x^2+x+1)^2} +\frac{1}{x-1}- \frac{1}{(x-1)^2}$ which is easy to integrate, even wolfram alpha don't show how to turn $ \frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2} $ into partial fractions and give a message "possible intermediate steps "

I want to know if there is an easy solution to this integral? I believe there must be an easy way to solve this without partial fractions.

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EDIT: here is my work

$$ \color{red}{\text{WARNING: THIS IS AN UGLY SOLUTION}}$$

$$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{a}{x-1}+ \frac{b}{(x-1)^2}+ \frac{cx+d}{x^2+x+1}+ \frac{ex+f}{(x^2+x+1)^2} $$

$b$ is easy to find (multiply both side by $(x-1)^2$ then substitute $x=1$ you will find $b=-1$ ) now I will try to find $a$

$$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{a}{x-1}- \frac{1}{(x-1)^2}+ \frac{cx+d}{x^2+x+1}+ \frac{ex+f}{(x^2+x+1)^2} $$ here I multiplied both sides by $(x-1)$ and then I added $\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^2+x+1)^2 (x-1) } + \frac{1}{(x-1)}$ and then $$\frac{3x^5 +4x^3-9x^2+2}{(x^2+x+1)^2 (x-1) } =a + (x-1) \left(\frac{cx+d}{x^2+x+1}+ \frac{ex+f}{(x^2+x+1)^2}\right)$$ then I used division to show $\frac{3x^5 +4x^3-9x^2+2}{(x^2+x+1)^2 (x-1) } = \frac{3x^4+3x^3+7x^2 -2x -2}{(x^2+x+1)^2}$ and I substituted $x=1$ and got $a=1$ so

$$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{1}{x-1}- \frac{1}{(x-1)^2}+ \frac{cx+d}{x^2+x+1}+ \frac{ex+f}{(x^2+x+1)^2} $$

here I used complex numbers let $w :=-\frac{1}{2} +\frac{\sqrt{3}}{2}i $ and $w^2 :=-\frac{1}{2} -\frac{\sqrt{3}}{2}i $ note that $w,w^2$ are roots for $x^2+x+1$ now $e,f$ are easy to find just like $b$ ie I multiplied both sides by $(x^2+x+1)^2$ and substituted $x=w$ after some calculation I found $4w^2+2w =ew^2+fw$ then $e=4 \ , \ f=2 $

$$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{1}{x-1}- \frac{1}{(x-1)^2}+ \frac{cx+d}{x^2+x+1}+ \frac{4x+2}{(x^2+x+1)^2} $$

for $c,d$ I would find then just like how I found $a$ i.e I multiplied multiply both sides by $x^2+x+1$ and add $\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x-1)^2 (x^2+x+1)} - \frac{4x+2}{(x^2+x+1)} $ and I used division to find that $\frac{3x^5 -x^4 -2x^3 -6x^2-2x+1}{(x-1)^2 (x^2+x+1)}= \frac{3x^3-4x^2-x-1}{(x-1)^2}$ then I substituted $x=w$ then $c=2 \ , \ d=1$

so $$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{1}{x-1}- \frac{1}{(x-1)^2}+ \frac{2x+1}{x^2+x+1}+ \frac{4x+2}{(x^2+x+1)^2} $$

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then $$ \int\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}dx=\ln({x-1})+ \frac{1}{(x-1)}+ \ln({x^2+x+1}) -2\frac{1}{(x^2+x+1)} +C $$

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EDIT

in the steps that I used division it would be better if I used l'hospital rule as $x \to k$ ($k=1$ when I wanted to find $a$ and $k=w$ when I wanted to find $c,d$) and rely on the fact that the function is continuous

Answer 1

I tried to remove one term such that $$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=\frac{Q(x)}{(x^3-1)^2}+\frac{P(x)}{(x^3-1)}$$ such that $P(x)$ be a quadratic in $x$ $$P(x)=a+b x+c x^2$$

Removing the denominator and grouping $$(-a-b c+1)-2 x-(c+12) x^2+x^3 (a+b c+2)-x^4+(c+3) x^5=0$$ So, $c=-3$ seems to be an idea.

Replacing $$(-a+3 b+1)-2 x-9 x^2+x^3 (a-3 b+2)-x^4=0$$ and we cannot do more. So

$$\frac{3x^5 -x^4 +2x^3 -12x^2-2x+1}{(x^3-1)^2}=-\frac{x^4-2 x^3+9 x^2+2 x-1}{\left(x^3-1\right)^2}-\color{red}{\frac{3x^2}{x^3-1}}$$

Now, partial fraction decomposition $$\frac{x^4-2 x^3+9 x^2+2 x-1}{\left(x^3-1\right)^2}=\frac{1}{(x-1)^2}-2\color{red}{\frac{2 x+1}{\left(x^2+x+1\right)^2}}$$ which is now very simple since $\cdots ???$ (fill the blanks).

Thsi was not a difficult integral.

Answer 2

By integration by parts $$ \begin{aligned} & \int \frac{3 x^5-x^4+2 x^3-12 x^2-2 x+1}{x^6-2 x^3+1} d x \\ = & \frac{1}{2} \int \frac{6 x^5-6 x^2}{x^6-2 x^3+1}+\frac{-x^4+2 x^3-9 x^2+1}{(x^3-1)^2} d x \\ = & \frac{1}{2} \ln \left(x^3-1\right)^2-\int \frac{-x^4+2 x^3-9 x^2+1}{3 x^2} d\left(\frac{1}{x^3-1}\right) \\ = & \ln \left|x^3-1\right|+\int\left(\frac{x^2}{3}-\frac{2}{3} x+3+\frac{2}{3x}-\frac{1}{3 x^2}\right) d\left(\frac{1}{x^3-1}\right) \\ = & \ln \left|x^3-1\right|+\frac{1}{x^3-1}\left(\frac{x^2}{3}-\frac{2}{3} x+3+\frac{2}{3x}-\frac{1}{3 x^2}\right) - \int \frac{1}{x^3-1}\left(\frac{2 x}{3}-\frac{2}{3}-\frac2{3x^2}+\frac{2}{3 x^3}\right) d x\\=& \ln \left|x^3-1\right|+\frac{1}{x^3-1}\left(\frac{x^2}{3}-\frac{2}{3} x+3+\frac2{3x}-\frac{1}{3 x^2}\right) -\frac23 \int\left(\frac{1}{x^2}-\frac{1}{x^3}\right) d x\\=& \ln \left|x^3-1\right|+ \frac{x^2-x+3}{x^3-1}+C \end{aligned} $$

Answer 3

Possibly a more detailed approach(?) that doesn't involve long-page calculation:

Firstly, observe that the coefficients of the terms $x ^ 4, x ^ 3, x$ and the constant term 1 seem to have some secrets hidden about them. So we'd like to separate them out, and it turns out this is indeed correct.

Secondly, notice that the remaining part $$\int \frac{3x ^ 5 - 12 x ^ 2}{(x ^ 3 - 1) ^ 2}$$ can be split into $$I = \int \frac{3x ^ 5 - 3 x ^ 2}{(x ^ 3 - 1) ^ 2}$$ and $$J = \int \frac{-9x ^ 2}{(x ^ 3 - 1) ^ 2},$$

both of which are able to be integrated (why???).

Then, about the secret of the symmetric part:

$$-x ^ 4 + 2x ^ 3 - 2x + 1 = (x - 1) ^ 2(-x ^ 2 + 1).$$ So this symmetric part of the integral becomes $$K = \int \frac{(x - 1) ^ 2 (-x ^ 2 + 1)}{(x ^ 3 + 1) ^ 2} = \int \frac{(x - 1) ^ 2 (-x ^ 2 + 1)}{(x - 1) ^ 2(x ^ 2 + x + 1) ^ 2} = \int \frac{-x ^ 2 + 1}{(x ^ 2 + x + 1) ^ 2}.$$

The following is in my opinion the key observation to make, and actually the most difficult observation: $$\frac{d}{dx}(\frac{x}{x ^ 2 + x + 1}) = \frac{(x ^ 2 + x + 1) - x(2x + 1)}{(x ^ 2 + x + 1) ^ 2} = \frac{-x ^ 2 + 1}{(x ^ 2 + x + 1) ^ 2},$$ from which

$$K = \frac{x}{x ^ 2 + x + 1} + C.$$

Answering the "(why???)" part: $$I = \frac{1}{2} \ln |(x ^ 3 - 1) ^ 2| = \ln |x ^ 3 - 1|,$$ and $$J = 3(x ^ 3 - 1) ^ {-1} = \frac{3}{x ^ 3 - 1}.$$

Summing up, the answer is $$I + J + K = \ln |x ^ 3 - 1| + \frac{3}{x ^ 3 - 1} + \frac{x}{x ^ 2 + x + 1} + C,$$ for some arbitrary constant C.