Resolving $\int \frac{1}{(x-1)^3-x}\mathrm{d}x$

Question

How can I resolve this primitive: $$\int \frac{1}{(x-1)^3-x}\mathrm{d}x$$

I was trying to use the partial fractions, but I don't know how to use it in this type of equation.

$$\int \frac{1}{(x-1)^3-x}\mathrm{d}x =\int \frac{1}{x^3-3x^2+2x-1}\mathrm{d}x$$

Answer 1

Hint

Consider $$x^3-3 x^2+2 x-1=0$$ and use Cardano method. You should notice that there is only one real root; just name it $\alpha$ (do not try to use its value; it is more than likely a monster). So, $$x^3-3 x^2+2 x-1=(x-\alpha)(x^2+a x+b)$$ You need to identify $a,b$ (which are functions of $\alpha$) in order to make things clear and clean later; but the imortant thing is that $x^2+ax+b$ does not show any real root.

Now, partial fraction decomposition will take you to two standard integrals and the end result would be a rather simplex expression in which all terms will be functions of $\alpha$, the real root of the initial equation.

I am sure that you can take it from here.

Answer 2

By using the substitution $x=y+1$ we just have to compute the anti-derivative of $\frac{1}{y^3-y-1}$.

Now $y^3-y-1$ has a unique real root $\alpha\in(1,2)$, hence two conjugated complex roots $\beta,\overline{\beta}$ fulfilling $\alpha+\beta+\overline{\beta}=0$ and $\alpha\beta\overline{\beta}=1$ by Viète's theorem. If $Z=\{\alpha,\beta,\overline{\beta}\}$, for any $z\in Z$:

$$ \text{Res}\left(\frac{1}{y^3-y-1},y=z\right) = \frac{1}{3z^2-1} $$ hence: $$ \int \frac{dy}{y^3-y-1} = C+\sum_{z\in Z}\frac{\log(y-z)}{3z^2-1}.$$

We may locate the roots of $p(y)=y^3-y-1$ also by noticing that $p(y)=0$ is equivalent to:

$$ T_3\left(\frac{\sqrt{3}}{2}\,y\right)=\frac{3\sqrt{3}}{2} $$ with $T_3$ being a Chebyshev polynomial of the first kind, so if $y=\frac{2}{\sqrt{3}}\cosh(\theta)$, $\cosh(3\theta)=\frac{3\sqrt{3}}{2}$, giving: $$\begin{eqnarray*} \alpha &=& \frac{2}{\sqrt{3}}\cosh\left(\frac{1}{3}\text{arccosh}\left(\frac{3\sqrt{3}}{2}\right)\right)\\&=&\frac{2}{\sqrt{3}}\cosh\left(\frac{1}{3}\log\left(\frac{\sqrt{23}+\sqrt{27}}{2}\right)\right)\\&=&\frac{1}{\sqrt{3}}\left(\sqrt[3]{\frac{\sqrt{27}+\sqrt{23}}{2}}+\sqrt[3]{\frac{\sqrt{27}-\sqrt{23}}{2}}\right) \\&=&1.324717957244746\ldots \end{eqnarray*}$$ and $$ \beta,\overline{\beta} = \frac{1}{\sqrt{3}}\left(\omega^{\pm 1}\sqrt[3]{\frac{\sqrt{27}+\sqrt{23}}{2}}+\omega^{\mp 1}\sqrt[3]{\frac{\sqrt{27}-\sqrt{23}}{2}}\right)$$ with $\omega=e^{\frac{2\pi i}{3}}=\frac{-1+i\sqrt{3}}{2}$.

From the decomposition $y^3-y-1 = (y-\alpha)(y^2-\alpha y-(\alpha^2-1)) $ we get:

$$ \int\frac{dy}{y^3-y-1} = -\frac{\log(y-\alpha)}{\alpha^2+1}-\frac{1}{1+\alpha^2}\int\frac{y\,dy}{1+\alpha^2+\alpha y-y^2}. $$